Chapter 20: Q70P (page 608)

A45.0 gblock of tungsten at $30.0 ° C$ and a 25.0 gblock of silver at $- 120.0 ° C$ are placed together in an insulated container. (See Table 18-3 for specific heats.) (a) What is the equilibrium temperature? What entropy changes do (b) the tungsten, (c) the silver, and (d) the tungsten–silver system undergo in reaching the equilibrium temperature?

Short Answer

Expert verified

a) The equilibrium temperature is $- 44.2 ° C$ .

b) The entropy change, that the tungsten undergoes in reaching the equilibrium temperature is -1.69 J/K.

c) The entropy change that the silver undergoes in reaching the equilibrium temperature is 2.38 J/K.

d) The entropy change, that the tungsten-silver system undergoes in reaching the equilibrium temperature is 0.69 J/K.

Step by step solution

The given data

a) Mass of the tungsten block, $m_{1} = 0.045 kg$

b) Mass of silver, $m_{2} = 0.025 kg$

c) Temperature of tungsten, $T_{1} = 30 ° C = 303 K$

d) Temperature of silver, $T_{2} = - 120 m ° C = 153 k$

e) Specific heat of tungsten, $c_{1} = 134 J / kg . K$

f) Specific heat of silver, $c_{2} = 236 J / kg . K$

Understanding the concept of thermal equilibrium and entropy change

Using the concept of thermal equilibrium, the net heat absorbed or released by the system can be calculated. When two objects are made in contact within an insulated container, they achieve a state of equilibrium at a given temperature. In this case, the heat released by the hot body is equal to the heat absorbed by the cold body.

From the second law of thermodynamics, the entropy change of a body for a reversible process can be given as the net heat released by the body to the system per the temperature of the system.

Formulae:

The heat absorbed or released by a body, $Q = m c ∆ T$ ……(i)

The entropy of a body, $∆ S = \int_{T_{1}}^{T_{2}} \frac{dQ}{T}$ ……(ii)

a) Calculation of the equilibrium temperature

Let the equilibrium temperature of the two materials be $T_{f}$ .

For the equilibrium state of the tungsten-silver system, the net heat absorbed or released by the system can be given and equilibrium temperature can be calculated as follows:

$\sum_{} Q = 0 Q_{t u n s t e n} + Q_{s i l v e r} = 0 m_{1} c_{1} (T_{f} - T_{1}) m_{2} c_{2} (T_{f} - T_{2}) = 0 (m_{1} c_{1} + m_{2} c_{2}) T_{f} = m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2} T_{f} = \frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2}}{(m_{1} c_{1} + m_{2} c_{2})} = \frac{(0.045 kg) (134 J / kg . K) (303 K) + (0.025 kg) ((236 J / kg . K)) (153 K)}{(0.045 kg) (134 J / kg . K) + (0.025 kg) ((236 J / kg . K))} = 228.8 K = - 44.2 ° C$

Hence, the equilibrium temperature is $- 44.2 ° C$ .

b) Calculation of the entropy change of tungsten

Substituting equation (i) in equation (ii), the entropy change of tungsten element within the insulated container can be given as follows:

$∆ S = \int_{T_{1}}^{T_{1}} \frac{m_{1} c_{1} d t}{T} = m_{1} c_{1} \int_{T_{1}}^{T_{1}} \frac{d T}{T} = m_{1} c_{1} I n {[T]}_{T_{1}}^{T_{1}} = m_{1} c_{1} I n [\frac{T_{f}}{T_{1}}]$

For the given values, we have-

$∆ S = (0.045 kg) (134 J / kg . K) In [\frac{229 K}{303 K}] = 1.69 J / K$

Hence, the value of entropy change is -1.69 J/K.

c) Calculation of the entropy change of silver

The entropy change of silver element within the insulted container can be given as follows:

$∆ S = \int_{T_{1}}^{T_{1}} \frac{m_{2} c_{2} d t}{T} = m_{2} c_{2} \int_{T_{1}}^{T_{1}} \frac{d T}{T} = m_{2} c_{2} I n {[T]}_{T_{1}}^{T_{1}} = m_{2} c_{2} I n [\frac{T_{f}}{T_{1}}]$