Question

One kg of water at 27\(^\circ\)C is brought in contact with a heat reservoir kept at 37\(^\circ\)C. Upon reaching thermal equilibrium, this mass of water is brought in contact with another heat reservoir kept at 47\(^\circ\)C. The final temperature of water is 47\(^\circ\)C. The change in entropy of the whole system in this entire process is \rule{1cm{0.15mm} cal/K. (up to two decimal places)
Take specific heat at constant pressure of water as 1 cal/(g K)}

Hint:

The entropy change of the "universe" or "whole system" in an irreversible process must always be positive. If you calculate a negative value, re-check your signs. The entropy of a body providing heat (like a reservoir) decreases, while the entropy of a body receiving heat increases. The total change is the sum of these, and the increase in the colder body's entropy is always larger in magnitude than the decrease in the hotter body's entropy.

Answer 1

Correct Answer: 1.03

Step 1: Understanding the Concept:
The problem asks for the total entropy change of the universe (the "whole system," which includes the water and the two reservoirs) for a two-step heating process. The total entropy change is the sum of the entropy changes of each component. Since the heating is irreversible (due to finite temperature differences), the total entropy change of the universe must be positive.
Step 2: Key Formula or Approach:
- Change in entropy for an object with changing temperature: \(\Delta S = \int_{T_i}^{T_f} \frac{dQ}{T} = mc \ln\left(\frac{T_f}{T_i}\right)\). - Change in entropy for a heat reservoir at constant temperature: \(\Delta S = \frac{Q}{T}\), where Q is the heat absorbed by the reservoir. If the reservoir gives up heat, Q is negative. - Total entropy change: \(\Delta S_{total} = \Delta S_{water} + \Delta S_{reservoirs}\).
Step 3: Detailed Explanation:
First, convert all temperatures to Kelvin:
\(T_{w, initial} = 27^\circ\text{C} = 300\) K
\(T_{R1} = 37^\circ\text{C} = 310\) K
\(T_{R2} = 47^\circ\text{C} = 320\) K
Mass of water \(m = 1\) kg = 1000 g. Specific heat \(c = 1\) cal/(g·K).
Process 1: Water heated from 300 K to 310 K by Reservoir 1.
- Heat absorbed by water: \(Q_1 = mc(T_{R1} - T_{w, initial}) = 1000 \times 1 \times (310 - 300) = 10000\) cal.
- Entropy change of water: \(\Delta S_{w1} = mc \ln\left(\frac{310}{300}\right) = 1000 \ln(31/30) \approx 1000 \times 0.03279 = 32.79\) cal/K.
- Entropy change of Reservoir 1 (loses heat): \(\Delta S_{R1} = \frac{-Q_1}{T_{R1}} = \frac{-10000}{310} \approx -32.26\) cal/K.
Process 2: Water heated from 310 K to 320 K by Reservoir 2.
- Heat absorbed by water: \(Q_2 = mc(T_{R2} - T_{R1}) = 1000 \times 1 \times (320 - 310) = 10000\) cal.
- Entropy change of water: \(\Delta S_{w2} = mc \ln\left(\frac{320}{310}\right) = 1000 \ln(32/31) \approx 1000 \times 0.03175 = 31.75\) cal/K.
- Entropy change of Reservoir 2 (loses heat): \(\Delta S_{R2} = \frac{-Q_2}{T_{R2}} = \frac{-10000}{320} = -31.25\) cal/K.
Total Entropy Change of the Whole System:
The total change is the sum of the entropy changes of all parts:
\[ \Delta S_{total} = \Delta S_{w1} + \Delta S_{w2} + \Delta S_{R1} + \Delta S_{R2} \] \[ \Delta S_{total} = (32.79 + 31.75) + (-32.26 - 31.25) \text{ cal/K} \] \[ \Delta S_{total} = 64.54 - 63.51 = 1.03 \text{ cal/K} \] \textit{Alternative Calculation:}
Total entropy change of water from 300 K to 320 K:
\(\Delta S_{water} = mc \ln(320/300) = 1000 \ln(32/30) \approx 64.54\) cal/K. Total entropy change of reservoirs:
\(\Delta S_{res} = \Delta S_{R1} + \Delta S_{R2} = -32.26 - 31.25 = -63.51\) cal/K. Total entropy change of the universe:
\(\Delta S_{total} = \Delta S_{water} + \Delta S_{res} = 64.54 - 63.51 = 1.03\) cal/K. Step 4: Final Answer:
The change in entropy of the whole system is 1.03 cal/K.