How much energy is needed to melt 15 grams of ice to water at #0^.C# ?

1 Answer
May 16, 2018

In Physics, when talking about energy in thermodynamics, we talk about heat.

Heat can be defined with the letter Q.

Where: #Q = m * C * DeltaT#
#m# == mass
#C# == specific heat
#DeltaT# == change in temperature

Specific heat is defined as the amount of heat that is needed to raise a 1kg mass by 1ºC/1K. Units #[J/(kg*ºC)]#

Liquid water has a very high Specific Heat Capacity. It is about 4180 #J/(kg*ºC)#.

To solve your problem, we will have to introduce a new concept, phase change . This happens when a substance goes through a change in state, in other words, it changes from liquid to solid , liquid to gas or gas to solid.

In your case, we are melting ice, which means the phase change will go from solid to liquid.

To calculate this you need a new concept, Specific Latent Heat (denoted with the letter L), which is the amount of heat that has to be supplied to a mass to change its phase. This can be calculated by:

#Q = m*L#

To solve your problem, we will have to break it into parts:

  1. Heat needed to melt ice at 15ºC to 0ºC (still in solid state)
  2. Heat needed to melt solid ice at 0ºC to 0ºC liquid.

Before we start:

  • Specific Heat of Ice (C)= 2108 J/(kg*ºC)
  • Latent Heat of Ice (L) = 333*10^3 J/kg


#DeltaQ_{Total} = m*C*DeltaT + m* L#
#DeltaQ_{Total} = DeltaQ_{Ice} + DeltaQ_{Fusion}#

#DeltaQ_{Ice} = ((0.015kg)(2108J/(kgºC))(0ºC - 15ºC))#
#DeltaQ_{Fusion} = ((0.015kg)(333*10^3))#

#DeltaQ_{Total} = ((0.015kg)(2108J/(kgºC))(0ºC - 15ºC)) + ((0.015kg)(333*10^3))#

#DeltaQ_{Total} = -473.3 J + 4995J#
#DeltaQ_{Total} = 4520.7J#

This means that to melt a 0.015kg ice cube to water at 0ºC you will need 4520.7 J of heat.