(a) How much heat (J) flows from 1.00 kg of water at 46.0°C when it is placed in contact with 1.00 kg of 18°C water in reaching equilibrium?
(b) What is the change(J/K) in entropy due to this heat transfer?
(c) How much work (J) is made unavailable, taking the lowest temperature to be 18°C?
To answer these questions, we can use the principles of thermodynamics.
(a) The amount of heat (Q) transferred can be determined using the equation:
Q = m * c * ΔT
Where:
Q is the heat transferred
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
For the first part, we have two cases: heating the water at 46°C and cooling the water at 18°C.
Case 1: Heating from 18°C to 46°C
m = 1.00 kg
c = specific heat capacity of water = 4.18 J/g°C
ΔT = 46°C - 18°C = 28°C
Q1 = m * c * ΔT
Q1 = 1.00 kg * 4.18 J/g°C * 28°C
Q1 = 1174.24 J
Case 2: Cooling from 46°C to 18°C (assuming no heat loss to the surroundings)
m = 1.00 kg
c = specific heat capacity of water = 4.18 J/g°C
ΔT = 46°C - 18°C = 28°C
Q2 = m * c * ΔT
Q2 = 1.00 kg * 4.18 J/g°C * 28°C
Q2 = 1174.24 J
Since the heat transfer during heating is equal to the heat transfer during cooling (assuming no heat loss), the total heat transferred is:
Q_total = Q1 + Q2
Q_total = 1174.24 J + 1174.24 J
Q_total = 2348.48 J
Therefore, the heat flow from the water at 46°C to the water at 18°C when reaching equilibrium is 2348.48 J.
(b) The change in entropy (ΔS) due to heat transfer can be calculated using the equation:
ΔS = Q / T
Where:
ΔS is the change in entropy
Q is the heat transferred (2348.48 J from part a)
T is the average temperature during heat transfer
To find the average temperature, we can assume that it is the average of the initial and final temperatures:
T_avg = (46°C + 18°C) / 2
T_avg = 32°C
ΔS = Q / T_avg
ΔS = 2348.48 J / 32°C
ΔS = 73.39 J/K
Therefore, the change in entropy due to this heat transfer is 73.39 J/K.
(c) To find the work made unavailable during this process, we need to determine the change in available energy (ΔU) and the lowest temperature involved.
The change in available energy is given by:
ΔU = Q - W
Where:
ΔU is the change in available energy
Q is the heat transferred (2348.48 J from part a)
W is the work done or made unavailable
To calculate the work, we can use the equation:
W = Q_lowest * (1 - T_lowest / T_highest)
Where:
Q_lowest is the heat transferred at the lowest temperature (Q2 = 1174.24 J from part a)
T_lowest is the lowest temperature (18°C)
T_highest is the highest temperature (46°C)
Using these values in the equation:
W = 1174.24 J * (1 - 18°C / 46°C)
W = 1174.24 J * (1 - 0.3913)
W = 709.9 J
Therefore, the amount of work made unavailable, taking the lowest temperature to be 18°C, is 709.9 J.