Question

How much heat is needed to melt 10.0 grams of ice at -10°C until it is water at 10°C?

Answer 1

Taking into account the definition of calorimetry, sensible heat and latent heat, the heat needed to melt 10.0 grams of ice at -10°C until it is water at 10°C is 3,969.5 J.

Calorimetry

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

In this case, the amount of heat a body receives or transmits is determined by:

Q = c× m× ΔT

where:

• Q is the heat exchanged by a body of mass m.
• c is a specific heat substance.
• ΔT is the temperature variation.

Latent heat

Latent heat is defined as the energy required by a quantity of substance to change state.

When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.

In this case, the heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to

Q = m×L

where L is called the latent heat of the substance and depends on the type of phase change.

This case

First, you  have to get the ice from -10°C to 0 °C, which is the melting point.  Then you have to melt the ice into liquid water. Now, you have to get the water from 0°C to 10°C.

-10°C to 0 °C

In this case, you know:

• c= specific heat capacity of ice= 2.108 $\frac{J}{gK}$
• m= 10 g
• ΔT= Tfinal - Tinitial= 0 °C - (-10 °C)= 10 °C= 10 K because being a temperature difference, the difference is the same in °C and K.

Replacing in the definition of sensible heat:

Q1= 2.108 $\frac{J}{gK}$× 10 g× 10 K

Solving:

Q1= 210.8 J

Heat needed to melt ice

In this case, you have to melt the ice into liquid water. Being the specific heat of melting of ice is 334 J/g, the heat needed to melt 10 grams of ice is calculated as:

Q2= 10 grams× 334 J/g

Solving:

Q2= 3,340 J

0°C to 10 °C

In this case, you know:

• c= specific heat capacity of liquid water is 4.187 $\frac{J}{gK}$
• m= 10 g
• ΔT= Tfinal - Tinitial= 10 °C - 0 °C= 10 °C= 10 K because being a temperature difference, the difference is the same in °C and K.

Replacing in the definition of sensible heat:

Q3= 4.187 $\frac{J}{gK}$× 10 g× 10 K

Solving:

Q3= 418.7J

Total heat required

The total heat required is calculated as:  

Total heat required= Q1 + Q2 + Q3

Total heat required= 210.8 J + 3,340 J + 418.7 J

Total heat required= 3,969.5 J

In summary, the heat needed to melt 10.0 grams of ice at -10°C until it is water at 10°C is 3,969.5 J.

Learn more about calorimetry:

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