How much heat is needed to convert a cube of ice with mass 20g and Temperature -20°C into water of 20°C?

Answer

You need three times energy for that: first to bring the ice to 0°C (Q1) , then to melt the ice in water at 0°C (Q2), and finally to bring water from 0°C to 20°C to bring (Q3).

The first and third step can be found with the formula Q = mcdT

with m : the mass of the substance to be heated in kg,

c : the specific heat capacity in Joule/kg/K (Joule per kilogram, per Kelvin) = the energy required to heat 1 kg by 1 degree (so with 1 Kelvin)

dT : the number of Kelvin by which you want to increase the temperature.

For ice you can easily find on the internet: c(ice) = 2108 J/(kg.K) and for water c(water) = 4187 J/(kg.K)

So: to bring the ice to 0°C: Q1 = 0.02 . 2108 . 20 = 843 J

and to bring the water from 0°C to 20°C: Q3 = 0.02 . 4187 . 20 = 1675 J

Then there is the intermediate step, melting ice in water. It takes 334 000 Joules to convert one kg of ice into water at 0°C.

So here, with 20 gr instead of 1 kg that is: Q2 = 0.02 . 334 000 = 6 680 J

Together: Q1 + Q2 + Q3 = 843 + 1675 + 6680 = 9 198 Joules

So two thirds goes to the melting…