Question

In reaching equilibrium, how much heat transfer occurs from 1.00 kg of water at 40.0º C when it is placed in contact with 1.00 kg of 20.0º 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 20.0º C? Explicitly show how you follow the steps in the Problem-Solving Strategies for Entropy.

a) 40,000 J, 160 J/K, 24,000 J
b) 45,000 J, 180 J/K, 28,000 J
c) 50,000 J, 200 J/K, 32,000 J
d) 55,000 J, 220 J/K, 36,000 J

Answer 1

To find the heat transfer, we can use the equation Q = mcΔT and calculate the change in entropy using ΔS = Q/T. The amount of work made unavailable can be determined using W = ΔS × T.

Step-by-step explanation:

To determine the heat transfer in this situation, 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 change in temperature.

Using this equation, we find that the heat transfer from the 1.00 kg of water at 40.0º C to the 1.00 kg of 20.0º C water is Q = (1.00 kg)(4186 J/kgºC)(20.0ºC - 40.0º C) = -83600 J. Note that the negative sign indicates that heat is being transferred out of the system.

The change in entropy due to this heat transfer can be calculated using the equation ΔS = Q/T, where ΔS is the change in entropy, Q is the heat transfer, and T is the temperature in Kelvin.

For the first step of the problem-solving strategy, we need to convert the temperatures to Kelvin. 40.0º C + 273 = 313 K and 20.0º C + 273 = 293 K. Plugging in the values, ΔS = -83600 J / 313 K = -267 J/K.

To calculate the amount of work made unavailable, we can use the equation W = ΔS × T. Plugging in the values, W = -267 J/K × 293 K = -78331 J.