In the house to rooms are separated by a wall made of brick 4 inch thick Room A is held a constant temperature a) At what rate is his transferred from room into room key while warranty is at the indicated temperature? b) find the change in entropy involved in bringing a room B to the temperature of the room A c) find a total change in entropy for the two rooms due to this process

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In the house to rooms are separated by a wall made of brick 4 inch thick Room A is held a constant temperature

a) At what rate is his transferred from room into room key while warranty is at the indicated temperature?

b) find the change in entropy involved in bringing a room B to the temperature of the room A

c) find a total change in entropy for the two rooms due to this process

A:
B:
Density of air: 1.29
kg/m^3
Spec heat of air:
1006 J/kgK
T= 300K
T= 273K
Volume: 60 m^3
Area of wall : 12m^2
Transcribed Image Text:A: B: Density of air: 1.29 kg/m^3 Spec heat of air: 1006 J/kgK T= 300K T= 273K Volume: 60 m^3 Area of wall : 12m^2
Expert Solution
Step 1

Temperature of room A (TA)=300 K

Area of wall in room A (AA)=12 m2

Temperature of room B (TB)=273 K

Volume of room B (TB)=60 m3

Density of air (ρ)=1.29 kg/m3

Specific heat of air (C)=1006 J kg-1 K-1

Thermal conductivity of brick (k)=1.7 W m-1 K-1

Thickness of the brick wall (x)=4 inch=0.1016 m

Step 2

As the temperature of room A is greater than temperature of room B, heat will transfer from room A to room B till they reach equilibrium.

Therefore, this rate of heat transfer is given by the equation;

dQdt=kATA-TBx

Substituting the values;

dQdt=1.7×12×(300-273)0.1016

dQdt=5.42 kW

 

Thus, the rate of heat transfer from room A to room B is 5.42 kW.

Step 3

To bring room B to the temperature of room A, the change in heat energy will be;

Q=nCTA-TB

Substituting the values;

Q=4×1006×(300-273)

Q=108648 J

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