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

May 16, 2018

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

Heat can be defined with the letter Q.

Where: $Q = m \cdot C \cdot \Delta T$
$m$ == mass
$C$ == specific heat
$\Delta T$ == 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:

$Q = m \cdot L$

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

Hence:

$\Delta {Q}_{T o t a l} = m \cdot C \cdot \Delta T + m \cdot L$
$\Delta {Q}_{T o t a l} = \Delta {Q}_{I c e} + \Delta {Q}_{F u s i o n}$

DeltaQ_{Ice} = ((0.015kg)(2108J/(kgºC))(0ºC - 15ºC))
$\Delta {Q}_{F u s i o n} = \left(\left(0.015 k g\right) \left(333 \cdot {10}^{3}\right)\right)$

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

$\Delta {Q}_{T o t a l} = - 473.3 J + 4995 J$
$\Delta {Q}_{T o t a l} = 4520.7 J$

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