A 3.8 kg block of copper at a temperature of 84°C is dropped into a bucket containing a mixture of ice and water whose total mass is 1.2 kg. When thermal equilibrium is reached the temperature of the water is 8°C. How much ice was in the bucket before the copper block was placed in it? (Neglect the heat capacity of the bucket.)

? kg

0.211 kg

Explanation:

specific heat capacity of copper = 385 J/Kgk

heat loss by copper = mcθ = 3.8 * (8 - 84) * 385 = - 111188 J

heat needed to raise the temperature of water from 0°C to 8°C

= mcθ = 1.2 kg * 4180 * (8 - 0) = 40128 J

111188 J - 40128 J = 71060

71060 = ml

71060 / 336000 = mass of ice where latent heat of fusion = 3.36 * 10⁵JKg⁻¹

m = 0.211 kg

0.215 kg of ice

Explanation:

Mass of copper, m = 3.8 kg

Specific heat of Copper, Cp = 385 J/kg•K;

Change in temperature = 84 - 8

= 76°C

q = m * Cp * delta T

= 385 * 3.8 * 76

= 111188 J

111188 J = 26.575 kcal

The energy needed to heat 1.2 kg of water from 0°C to 8°C = q

q = m * Cp * delta T

= (1.2 kg) (1 kcal/kg•°C) (8°C)

= 9.6 kcal

Therefore,

delta q = q1 - q2

= 26.575 - 9.6

= 16.975 kcal were used to melt the ice.

Energy needed to melt ice at 0°C = 79 kcal/kg

Mass = 16.975/79

= 0.215 kg of ice.

14 kcal / (79 kcal/kg) = 0.18 kg.