Question

What mass of ice at -10.5°C is needed to cool a whale's water tank, holding 1.13 103 m3 of water, from 20.0°C down to a more comfortable 10.0°C?

Answer 1

SOLUTION:

We know that the heat lost by the water is the heat gained by ice.

Given that:

Initial temperature of ice,

Initial temperature of water

Final temperature of the mixture

Mass of water, m = Volume x density = 1.13 x 10³ m³ x 1000 kg/m³ = 1.13 x 10⁶ kg

Heat capacity of water, C = 4184 J/kg^oC

Heat capacity of ice, C_ice = 2093 J/kg^oC

Latent heat of fusion of ice, L_f = 343 x 10³​​​​​​​ J/kg

Heat lost by ice = heat gained by water

Heat lost by ice is given by: (Heat required to rise temp to 0^oC (melting point) + heat required for phase change (Lf.m) + heat required to raise the liquid water from 0 to 10)

heat gained by water is given by:

We have,

That is 118846.8 kg of ice is required to cool the water in tank.