10 kg of ice at −10°C is mixed with 100 kg of water at 25°C.

Given:

• S_ice = 0.5 cal g^-1 °C^-1
• S_water = 1 cal g^-1 °C^-1
• L = 80 cal g^-1

Find the equilibrium temperature of the mixture.

Solution:

Convert masses:

Ice = 10 kg = 10000 g

Water = 100 kg = 100000 g

Heat required by ice:

Q₁ = 10000 × 0.5 × 10

= 50000 cal

Heat for melting:

Q₂ = 10000 × 80

= 800000 cal

Heat to raise melted water from 0°C to T:

Q₃ = 10000T

Total heat gained:

Q_gain = 850000 + 10000T

Heat lost by water:

Q_loss = 100000(25−T)

At equilibrium:

100000(25−T) = 850000 + 10000T

2500000 − 100000T = 850000 + 10000T

1650000 = 110000T

T = 15°C

Answer:

15°C