(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?
456
To solve these questions, we need to apply some concepts from thermodynamics. Let's break down each question step by step:
(a) To find the amount of heat that flows from one substance to another, we can use the formula:
Q = mcΔT
where:
Q represents the amount of heat transferred,
m is the mass of the substance,
c is the specific heat capacity of the substance, and
ΔT is the change in temperature.
In this case, we have two substances, both water, and we need to calculate the heat transferred when they reach equilibrium. We can assume that there is no heat lost to the surroundings during this process.
Given:
Mass of the first water sample (m1) = 1.00 kg
Initial temperature of the first water sample (T1) = 46.0°C
Mass of the second water sample (m2) = 1.00 kg
Initial temperature of the second water sample (T2) = 18°C
First, let's calculate the heat transferred when m1 cools down to the equilibrium temperature:
Q1 = m1cΔT1 = m1c(Teq - T1)
Next, let's calculate the heat transferred when m2 warms up to the equilibrium temperature:
Q2 = m2cΔT2 = m2c(Teq - T2)
Since the two substances reach equilibrium, we know that Q1 = -Q2.
Thus, the total heat transferred is:
Q = Q1 + Q2 = m1c(Teq - T1) + m2c(Teq - T2)
Substituting the given values into the equation will give you the answer.
(b) To find the change in entropy due to the heat transfer, we can use the formula:
ΔS = Q / T
where:
ΔS represents the change in entropy,
Q is the amount of heat transferred, and
T is the temperature at which the heat transfer occurs.
In this case, the heat transfer occurs between two substances at different temperatures. We can assume that the heat transfer is reversible, meaning that the temperature remains constant during the process.
Using the value of Q from part (a), and assuming that the entire heat transfer takes place at the temperature of the equilibrium, substitute the values into the formula to calculate the change in entropy.
(c) To determine the amount of work made unavailable, we need to use the concept of Carnot efficiency and the Second Law of Thermodynamics.
The Carnot efficiency is given by:
η = (T_high - T_low) / T_high
where:
η represents the Carnot efficiency,
T_high is the highest temperature, and
T_low is the lowest temperature.
In this case, the lowest temperature (T_low) is given as 18°C, and we can assume the highest temperature (T_high) to be the equilibrium temperature.
The work made unavailable (W_unavailable) is:
W_unavailable = Q * (1 - η)
Using the value of Q from part (a) and the value of η calculated using the Carnot efficiency formula, substitute the values to obtain the solution.
By following these steps, you should be able to find the answers to the questions.