Question

On a hot summer day, 4.00 × 10^6 J of heat transfer into a parked car takes place, increasing its temperature from 35.0º C to 45.0º C. What is the increase in entropy of the car due to this heat transfer alone?

a) 250 J/K
b) 300 J/K
c) 350 J/K
d) 400 J/K

Answer 1

To find the increase in entropy in the car when 4.00 × 10^6 J of heat is transferred and the temperature rises from 35.0°C to 45.0°C, one would typically divide the heat by the average temperature in Kelvin. However, the calculation reveals that none of the provided options match, suggesting a potential issue with the problem or choices.

Step-by-step explanation:

To calculate the increase in entropy of the car due to heat transfer we can use the formula:

ΔS = Q / T

where ΔS is the change in entropy, Q is the heat added to the system, and T is the temperature of the system. However, since the temperature changes, we must integrate over the temperature range to find the total entropy change:

ΔS = ∫(Q / T)dT

Because the temperature change is linear, we can approximate this integral by using the average temperature over the range 35.0°C to 45.0°C which is (35 + 45)/2 = 40.0°C or 313.15K. So the increase in entropy is:

ΔS = 4.00 × 10^6 J / 313.15 K

ΔS = (4.00 × 10^6 J) / (313.15 K) ≈ 12775 J/K

However, none of the provided options for ΔS are correct based on this calculation. Therefore, we either need more information or a reassessment of the problem setup and provided answer choices.