Chapter U5, Problem SI6RE

(a)

Interpretation Introduction

Interpretation: The final temperature of the water in the beaker to be calculated.

Concept Introduction: Thermodynamics is the analysis of heat transfer. Heat transfer direction is always from a hotter substance to colder one.

Answer to Problem SI6RE

The final temperature of the water in the beaker is $35°\text{C}$ .

Explanation of Solution

Given

  $\begin{array}{c}\text{mass of water }\left(\text{m}\right)=\text{ 1}00\text{ mL}\\ \text{=100}\cancel{\text{mL}}\text{×}\frac{\text{g}}{\cancel{\text{mL}}}\\ =100\text{g}\end{array}$

Temperature change $\left(\Delta \text{T}\right)=\text{T}-25°\text{C}$

Absorbed heat (H) =1000cal

Specific heat of water is $\text{1}\text{cal}/\text{g}\cdot °\text{C}$

The equation of heat is as follows:

  $\text{H=mCΔT}$

Here H is heat, m is mass, C is specific heat and T is temperature.

Put the values in above equation is as follows:

  $\begin{array}{c}\text{H=100}\cancel{\text{g}}\times \text{1}\frac{\text{cal}}{\cancel{\text{g}}\text{×}°\text{C}}\times \left(\text{T}-25\text{°C}\right)\\ \text{1000}\text{cal=100}\text{cal}/\text{°C}\times \left(\text{T}-25\text{°C}\right)\\ \text{T}-25\text{°C}=\text{10°C}\\ \text{T}=35\text{°C}\end{array}$

Thus, the final temperature of the water in the beaker is $35°\text{C}$ .

(b)

Interpretation Introduction

Interpretation: The final temperature of the coin to be calculated.

Concept Introduction: Thermodynamics is the analysis of heat transfer. Heat transfer direction is always from a hotter substance to colder one.

Answer to Problem SI6RE

The temperature of the coin is $35°\text{C}$ .

Explanation of Solution

Water and coin are reached in thermal equilibrium when they are put together in a beaker. After exchanging same amount of heat with each other both have same temperature. Thus the temperature of the coin is $35°\text{C}$ .

(c)

Interpretation Introduction

Interpretation: The specific heat of copper coin to be calculated.

Concept Introduction: Thermodynamics is the analysis of heat transfer. Heat transfer direction is always from a hotter substance to colder one.

Answer to Problem SI6RE

Specific heat of copper coin is $0.090\text{1}\text{cal}/\text{g°C}$ .

Explanation of Solution

Given mass of copper coin (m) = 30.4g

Temperature change

  $\begin{array}{c}\left(\Delta \text{T}\right)=400°\text{C}-35°\text{C}\\ \text{=}\text{365}°\text{C}\end{array}$

Absorbed heat (H) =1000cal

The equation of heat is as follows:

  $\text{H=mCΔT}$

Here H is heat, m is mass, C is specific heat and T is temperature.

Put the values in above equation is as follows:

  $\begin{array}{c}\text{1000}\text{cal=30}\text{.4g}\times \text{C}\times 365\text{°C}\\ \text{C=}\frac{\text{1000}\text{cal}}{\text{30}\text{.4g}\times 365\text{°C}}\\ \text{C}=\frac{\text{1000}\text{cal}}{\text{11096g}\times \text{°C}}\\ \text{T}=0.090\text{1}\text{cal}/\text{g°C}\end{array}$

Thus the specific heat of copper coin is $0.090\text{1}\text{cal}/\text{g°C}$ .