Two samples of water are mixed together.

The first sample has a mass of 0.50 kg and is at 0°C. The second sample has a mass of 1.5 kg and is at 100°C. What is the equilibrium temperature of the water, assume this is a closed system? (Cwater = 4.18 kJ/kg*°C.)

A. 75°C

B. 67°C

C. 33°C

D. 50°C

Answer;

A. 75°C

Explanation;

Let the change in temp of cold water be x degrees,

while that of hot water be 100 - x degrees.

Heat exchange = mcΔt

Ice

Δt = x

m = 0.50 kg

c = 4.18 kJ/kg*°C

Hot water

Δt = 100 - x

m = 1.5 kg

c = 4.18

But;

Heat lost = heat gained

0.50 * c * x = 1.5 * c * (100 - x)

0.50 * x = 1.5 * (100 - x)

0.5x = 150 - 1.5x

0.5x + 1.5x = 150 - 1.5x + 1.5x

2x = 150

x = 75° C

Hence; the equilbrium temperature will be 75° C