(a) How much heat 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 in entropy due to this heat transfer?
(c) How much work is made unavailable, taking the lowest temperature to be 18°C?
To answer these questions, we can use the principles of thermodynamics. Specifically, we can apply the concepts of heat transfer, entropy, and work to determine the values requested.
(a) To find out how much heat 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, we need to use the equation for heat transfer:
Q = mcΔT
Where:
Q is the heat transferred
m is the mass of the substance
c is the specific heat capacity (in this case, water's specific heat capacity is approximately 4186 J/(kg·°C))
ΔT is the change in temperature
By using this equation, we can calculate the heat transfer by finding the individual heat transfers for each water sample and then adding them together:
Q1 = m1cΔT1
Q2 = m2cΔT2
Where:
m1 and m2 are the masses of water samples 1 and 2 (both equal to 1.00 kg)
ΔT1 is the change in temperature for water sample 1 (T1 final - T1 initial)
ΔT2 is the change in temperature for water sample 2 (T2 final - T2 initial)
Finally, we sum the two heat transfers:
Q = Q1 + Q2
(b) To find the change in entropy (ΔS) due to heat transfer, we can use the equation:
ΔS = Q / T
Where:
ΔS is the change in entropy
Q is the heat transferred
T is the temperature in Kelvin (K)
We can convert the temperatures from Celsius to Kelvin by adding 273.15. To find the change in entropy, we divide the heat transfer by the final temperature (as the entropy change depends on the temperature at which it occurs).
(c) To find the amount of work made unavailable, we need to use the equation:
W_unavailable = Q / T_low
Where:
W_unavailable is the work made unavailable
Q is the heat transferred
T_low is the lowest temperature in Kelvin (K)
In this case, we take the lowest temperature to be 18°C, so we convert it to Kelvin by adding 273.15, and then we divide the heat transfer by this temperature.
By applying the appropriate equations and calculations, you can find the answers to the questions (a), (b), and (c) relating to heat transfer, change in entropy, and work made unavailable in this thermodynamic process.