Question

A copper vessel of mass 100 g contains 150 g of water at 50°C. How much ice is needed to cool it to 5°C?

Given: Specific heat capacity of copper = 0.4 Jg^-1 °C^-1

The Specific heat capacity of water = 4.2 Jg^-1 °C^-1

The Specific latent heat of fusion ice = 336 Jg^-1

Answer 1

Heat energy lost by the vessel and water contained in it in cooling the water from 50°C to 5°C is used in heating ice to melt it and then to raise its temperature from 0°C to 5°C.

Now, heat energy lost by the copper vessel is

Q_c = m_cc_cΔt  = 100 x 0.4 x (50 - 5)

Q_c = 1800 J

Similarly, heat energy lost by water is

Q_w = m_wc_wΔt = 150 x 4.2 x (50 - 5)

Q_w = 28350 J

Hence, the total heat energy lost is

Q_L = 1800 + 28350 = 30150 J

Let m g of ice be used to cool water. So, heat gained by ice is

Q_I = mL_ice + mc_wΔt

Q_I = 336m + m x 4.2 x 5 = 336m + 21m = 357mJ

Therefore, from the principle of calorimetry, the mass of ice is

Q_L = Q_I 

∴ 357m = 30150

∴ m = `30150/357` = 84.45 g