You have a reservoir held at a constant temperature of –30°C. You add 400 J of heat to the reservoir. If you have another reservoir at 0°C, how much heat must you add to it so that both reservoirs have the same increase in entropy?

449.38 J

Explanation:

ΔS = ΔQ/T

Where ΔS = entropy change

Q = quantity of heat

T = temperature

First reservoir :

T = –30°C = - 30 + 273 = 243K

Q = 400 J

Second reservoir :

T = 0°C = 273K

Q =?

To have same increase in entropy for both reservoirs :

Q/T of first reservoir = Q/T of second reservoir

400/243 = Q/273

243 * Q = 400 * 273

Q = (400 * 273) / 243

Q = 109,200 / 243

Q = 449.38271

Q = 449.38 J

b) 10 minutes

girls stopped for 10 minutes.