The temperatures of the stainless steel slab and copper slab from the surface.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 18, Problem 77P

To determine

The temperatures of the stainless steel slab and copper slab from the surface.

Expert Solution & Answer

Explanation of Solution

Given:

Initial uniform temperature (Ti) is 20°C.

Depth from the surface (x) is 1 cm.

Uniform heat flux (q˙s) is 8 kW/m2.

Total exposure time (t) is 60 s.

Calculation:

Write the properties of the stainless steel.

k=14.9 W/m⋅Kα=3.95×10−6 m2/s

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(1 cm)24(3.95×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc(0.3248)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.3248.

erfc(0.3248)=0.646

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)(0.646)]=20°C+4.9°C=24.9°C

Thus, the temperature of the stainless steel slab from the surface is 24.9°C.

Write the properties of the copper.

k=401 W/m⋅Kα=117×10−6 m2/s

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(1 cm)24(117×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc(0.05968)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.05968.

erfc(0.05968)=0.9327

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)(0.9327)]=20°C+1.7°C=21.7°C

Thus, the temperature of the copper slab from the surface is 21.7°C.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

(13) A cylindrical beam of length 2 m and diameter of 120 mm, is arranged with a loading in the middle and two supports either end, as shown in Figure Q13. Given the shaft is made of metal which has a tensile strength of 350 MPa. Select the safest Factor of Safety (FOS) to 1 decimal place that the design engineer should work to. 100 kN ○ A. 1.2 ○ B. No Valid Answer ○ c. 1.1 O D.3.7 E. 0.8 2 m Figure Q13 120 mm

The pin-connected assembly consists of bronze rods (1) and (2) and steel rod (3). The bronze rods each have a diameter of 12mm and an elastic modulus of E=120GPa. The steel rod has a diameter of 18mm and an elastic modulus of E 210 GPa. Assume a= 2.0 m, b=1.5 m, and c = 2.0 m. What is the magnitude of load P that is necessary to displace point A 5mm to the left? a 5.52kN 17.05kN 5.05kN d 6.75kN Right answer need, no chatgpt,only handwritten

The pin-connected assembly consists of bronze rods (1) and (2) and steel rod (3). The bronze rods each have a diameter of 12mm and an elastic modulus of E=120GPa. The steel rod has a diameter of 18mm and an elastic modulus of E 210 GPa. Assume a= 2.0 m. b = 1.5 m, and c = 2.0 m. What is the magnitude of load P that is necessary to displace point A 5 mm to the left? Solution must be in handwritten format,No Chatgpt 10 5.52kN b 17.05kN 5.05kN 6.75kN (3)

Answer 2

Question

Chapter 18, Problem 77P

To determine

The temperatures of the stainless steel slab and copper slab from the surface.

Expert Solution & Answer

Explanation of Solution

Given:

Initial uniform temperature (Ti) is 20°C.

Depth from the surface (x) is 1 cm.

Uniform heat flux (q˙s) is 8 kW/m2.

Total exposure time (t) is 60 s.

Calculation:

Write the properties of the stainless steel.

k=14.9 W/m⋅Kα=3.95×10−6 m2/s

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(1 cm)24(3.95×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc(0.3248)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.3248.

erfc(0.3248)=0.646

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)(0.646)]=20°C+4.9°C=24.9°C

Thus, the temperature of the stainless steel slab from the surface is 24.9°C.

Write the properties of the copper.

k=401 W/m⋅Kα=117×10−6 m2/s

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(1 cm)24(117×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc(0.05968)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.05968.

erfc(0.05968)=0.9327

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)(0.9327)]=20°C+1.7°C=21.7°C

Thus, the temperature of the copper slab from the surface is 21.7°C.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 18, Problem 77P

To determine

The temperatures of the stainless steel slab and copper slab from the surface.

Expert Solution & Answer

Explanation of Solution

Given:

Initial uniform temperature (Ti) is 20°C.

Depth from the surface (x) is 1 cm.

Uniform heat flux (q˙s) is 8 kW/m2.

Total exposure time (t) is 60 s.

Calculation:

Write the properties of the stainless steel.

k=14.9 W/m⋅Kα=3.95×10−6 m2/s

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(1 cm)24(3.95×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc(0.3248)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.3248.

erfc(0.3248)=0.646

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)(0.646)]=20°C+4.9°C=24.9°C

Thus, the temperature of the stainless steel slab from the surface is 24.9°C.

Write the properties of the copper.

k=401 W/m⋅Kα=117×10−6 m2/s

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(1 cm)24(117×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc(0.05968)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.05968.

erfc(0.05968)=0.9327

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)(0.9327)]=20°C+1.7°C=21.7°C

Thus, the temperature of the copper slab from the surface is 21.7°C.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 18, Problem 77P

To determine

The temperatures of the stainless steel slab and copper slab from the surface.

Expert Solution & Answer

Explanation of Solution

Given:

Initial uniform temperature (Ti) is 20°C.

Depth from the surface (x) is 1 cm.

Uniform heat flux (q˙s) is 8 kW/m2.

Total exposure time (t) is 60 s.

Calculation:

Write the properties of the stainless steel.

k=14.9 W/m⋅Kα=3.95×10−6 m2/s

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(1 cm)24(3.95×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc(0.3248)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.3248.

erfc(0.3248)=0.646

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)(0.646)]=20°C+4.9°C=24.9°C

Thus, the temperature of the stainless steel slab from the surface is 24.9°C.

Write the properties of the copper.

k=401 W/m⋅Kα=117×10−6 m2/s

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(1 cm)24(117×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc(0.05968)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.05968.

erfc(0.05968)=0.9327

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)(0.9327)]=20°C+1.7°C=21.7°C

Thus, the temperature of the copper slab from the surface is 21.7°C.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 18, Problem 77P

To determine

The temperatures of the stainless steel slab and copper slab from the surface.

Expert Solution & Answer

Explanation of Solution

Given:

Initial uniform temperature (Ti) is 20°C.

Depth from the surface (x) is 1 cm.

Uniform heat flux (q˙s) is 8 kW/m2.

Total exposure time (t) is 60 s.

Calculation:

Write the properties of the stainless steel.

k=14.9 W/m⋅Kα=3.95×10−6 m2/s

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(1 cm)24(3.95×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc(0.3248)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.3248.

erfc(0.3248)=0.646

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)(0.646)]=20°C+4.9°C=24.9°C

Thus, the temperature of the stainless steel slab from the surface is 24.9°C.

Write the properties of the copper.

k=401 W/m⋅Kα=117×10−6 m2/s

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(1 cm)24(117×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc(0.05968)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.05968.

erfc(0.05968)=0.9327

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)(0.9327)]=20°C+1.7°C=21.7°C

Thus, the temperature of the copper slab from the surface is 21.7°C.

Answer 6

Chapter 18, Problem 77P

To determine

The temperatures of the stainless steel slab and copper slab from the surface.

Expert Solution & Answer

Explanation of Solution

Given:

Initial uniform temperature (Ti) is 20°C.

Depth from the surface (x) is 1 cm.

Uniform heat flux (q˙s) is 8 kW/m2.

Total exposure time (t) is 60 s.

Calculation:

Write the properties of the stainless steel.

k=14.9 W/m⋅Kα=3.95×10−6 m2/s

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(1 cm)24(3.95×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc(0.3248)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.3248.

erfc(0.3248)=0.646

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)(0.646)]=20°C+4.9°C=24.9°C

Thus, the temperature of the stainless steel slab from the surface is 24.9°C.

Write the properties of the copper.

k=401 W/m⋅Kα=117×10−6 m2/s

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(1 cm)24(117×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc(0.05968)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.05968.

erfc(0.05968)=0.9327

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)(0.9327)]=20°C+1.7°C=21.7°C

Thus, the temperature of the copper slab from the surface is 21.7°C.

Answer 7

Chapter 18, Problem 77P

To determine

The temperatures of the stainless steel slab and copper slab from the surface.

Answer 8

Expert Solution & Answer

Explanation of Solution

Given:

Initial uniform temperature (Ti) is 20°C.

Depth from the surface (x) is 1 cm.

Uniform heat flux (q˙s) is 8 kW/m2.

Total exposure time (t) is 60 s.

Calculation:

Write the properties of the stainless steel.

k=14.9 W/m⋅Kα=3.95×10−6 m2/s

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(1 cm)24(3.95×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(3.95×10−6 m2/s)(60 s))]=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)erfc(0.3248)]

Refer Table 18-4, “The complementary error function”, obtain the value corresponding to the error function of 0.3248.

erfc(0.3248)=0.646

Calculate the temperature of the stainless steel slab from the surface (T(x,t)).

T(x,t)=20°C+8000 W/m214.9 W/m⋅K[4(3.95×10−6 m2/s)(60 s)πexp(−(0.01 m)24(3.95×10−6 m2/s)(60 s))−(0.01 m)(0.646)]=20°C+4.9°C=24.9°C

Thus, the temperature of the stainless steel slab from the surface is 24.9°C.

Write the properties of the copper.

k=401 W/m⋅Kα=117×10−6 m2/s

Calculate the temperature of the copper slab from the surface (T(x,t)).

T(x,t)=Ti+q˙sk[4αtπexp(−x24αt)−xerfc(x2αt)]=20°C+8 kW/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(1 cm)24(117×10−6 m2/s)(60 s))−(1 cm)erfc((1 cm)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc((0.01 m)2(117×10−6 m2/s)(60 s))]=20°C+8000 W/m2401 W/m⋅K[4(117×10−6 m2/s)(60 s)πexp(−(0.01 m)24(117×10−6 m2/s)(60 s))−(0.01 m)erfc(0.05968)]