Question

A thermos bottle contains 3.0kg of water and 2.0kg of ice in thermal equilibrium at 0ºC.How much heat is required to bring the system to thermal equilibriumat 50ºC?

Answer 1

First, the ice at 0 degrees Celsius must be transformed into the water at 0 degrees Celsius. The heat absorbed at this time is the so-called latent heat of fusion.

Q1 = quality * fusion latent heat of the water

Q1 = 2 * 334 = 668 KJ {latent heat of fusion of water = 334 KJ / Kg}

Now, there is 5 kg of water at 0 degrees Celsius, and the temperature can be increased to 50 degrees to absorb heat.

Q2 = specific heat of water * quality * (final temperature-initial temperature) {specific heat of water = 4.187 KJ / Kg}

Q2 = 4.187 * 5 * (50-0)

Q2 = 1046.75 KJ

Net heat required = Q2 + Q1 = 668 + 1046.75 = 1714.75 KJ

Answer = 1714.75 KJ