(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 the questions, we need to use the equations of heat transfer and entropy.
(a) To find the heat flow in joules (J) during the process, we can use the equation:
Q = mcΔT
where Q is the heat transfer, m is the mass, c is the specific heat capacity, and ΔT is the temperature change.
For water, the specific heat capacity (c) is approximately 4186 J/kg°C.
Given that the initial temperature of 1 kg of water is 46.0°C and the final temperature after reaching equilibrium with 1 kg of water at 18°C, ΔT = 46.0°C - 18°C = 28.0°C.
Substituting the values into the equation:
Q = (1.00 kg) * (4186 J/kg°C) * (28.0°C)
Q = 117,208 J
Therefore, 117,208 J of heat flows from the 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) The change in entropy (ΔS) is given by:
ΔS = Q / T
where ΔS is the change in entropy, Q is the heat transfer, and T is the temperature in Kelvin.
To calculate ΔS, we need to convert the temperatures from Celsius to Kelvin.
Initial temperature (Ti) = 46.0°C + 273.15 K = 319.15 K
Final temperature (Tf) = 18°C + 273.15 K = 291.15 K
Substituting the values into the equation:
ΔS = (117,208 J) / (319.15 K)
ΔS = 367.33 J/K
Therefore, the change in entropy due to this heat transfer is 367.33 J/K.
(c) The work made unavailable (Wunavailable) can be calculated using the equation:
Wunavailable = Qh - Qc
where Qh is the heat transferred from the hot reservoir (initially at 46.0°C) and Qc is the heat transferred to the cold reservoir (initially at 18°C).
Considering that the lowest temperature is 18°C and that Qc is the heat transferred to the cold reservoir:
Qc = mcΔTc
where mc is the mass of the cold water and ΔTc is the temperature change of the cold water.
Substituting the values:
Qc = (1.00 kg) * (4186 J/kg°C) * (28.0°C)
Qc = 117,208 J
Now, substituting the values into the equation for Wunavailable:
Wunavailable = (117,208 J) - (117,208 J)
Wunavailable = 0 J
Therefore, no work is made unavailable, as the heat transfer is entirely due to a temperature change and not a change in volume or pressure.