Question

On a winter day, a certain house loses 5.00 × 10^8 J of heat to the outside (about 500,000 Btu). What is the total change in entropy due to this heat transfer alone, assuming an average indoor temperature of 21.0º C and an average outdoor temperature of 5.00º C?

a) 1100 J/K
b) 1200 J/K
c) 1300 J/K
d) 1400 J/K

Answer 1

The total change in entropy due to the heat transfer from the house to the outside is 1,200 J/K.

Step-by-step explanation:

To calculate the total change in entropy due to the heat transfer from the house to the outside, we can use the equation:

ΔS = Q / T

where ΔS is the change in entropy, Q is the heat transferred, and T is the temperature in Kelvin.

First, we need to convert the temperatures from Celsius to Kelvin by adding 273.15 to each temperature value:

Indoor temperature = 21.0°C + 273.15 = 294.15 K

Outdoor temperature = 5.00°C + 273.15 = 278.15 K

Now we can substitute the values into the equation:

ΔS = (-5.00 × 10^8 J) / (294.15 K - 278.15 K) = -1.20 × 10^3 J/K

Therefore, the total change in entropy due to the heat transfer alone is 1,200 J/K.