how much energy is required to turn 20 gram cube of ice at a temperature of - 10 degrees Celsius into liquid water at a temperature of 10 degrees Celsius

Given that:

Mass (m) = 20 g = 0.02 Kg,

temperature (T₁) = - 10°C = - 10+273 = 263 K

temperature (T₂) = 10°C = 10+273 = 283 K

Specific heat of water (Cp) = 4.187 KJ/Kg k

We know that Heat transfer (Q) = m. Cp. (T₂ - T₁)

= 0.02 * 4.187 * (283-263)

Q = 1.67 KJ

Heat transferred is 1.67 KJ