Question

A copper calorimeter can with mass 0.860 kgcontains 0.185 kg of water and 0.020 kg of ice in thermal equilibrium at atmospheric pressure.

If 0.775 kg of lead at a temperature of 255 ∘C is dropped into the can, what is the final temperature of the system? (Assume no heat is lost to the surroundings.)

Express your answer in degrees Celsius to three significant figures.

Answer 1

The specific Heats are given by

Copper (solid) 0.385 (Units are kJ/kg.°K)
Lead (solid) 0.127
Water (liquid/25 °C) 4.181

And heat of fusion of ice = 333.6 kJ/kg

Now, let the final temperature be T °C. You can just use °C because you are concerned only with temperature differences and a Celsius degree is the same as Kelvin degree.

Since ice and water are in thermal equilibrium, the temperature must be 0°C

Heat lost by lead = heat gained by calorimeter, ice and water

Heat lost by lead = 0.775 x 0.127 x (255-T) kJ

Heat to melt ice at 0°C = 0.02 x 333.6 kJ
Heat to raise water to T°C from 0°C = 4.181T (0.185 + 0.02) kJ
Heat to raise calorimeter to T°C = 0.385*0.860T kJ

Therefore:

0.775 x 0.127 x (255-T) kJ= 0.02 x 333.6 kJ + 4.181T (0.185 + 0.02) kJ +0.385*0.860T

25.098-0.1115 T = 6.672 + 0.8571 T + 0.3311T

T = 14.177 ~ 14.2 degrees celsius