Question

The specific heat of solid water (ice) and liquid water are 2.03 J/gºC and 4.18 J/gºC respectively. Heating a 49.3 g sample of g ice from -25.0 °C to water at 44.7 °C requires 28.23 kJ of heat. Calculate the heat of fusion of water in J/g. Assume the melting point of water is 0 °C.​

Answer 1

Taking into account the definition of calorimetry, sensible heat and latent heat,  the heat of fusion of water is 335.091 $\frac{J}{g}$ .

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

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

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:

• -25 °C to 0 °C

In firts place, you know that the melting point of water is 0°C. So, first of all you must increase the temperature from -25° C (in solid state) to 0 ° C, in order to supply heat without changing state (sensible heat).

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, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

• c(solid)= 2.03 $\frac{J}{gC}$  
• m= 49.3 g
• ΔT= Tfinal - Tinitial= 0 °C - (-25 °C)= 25 °C

Replacing:

Q1= 2.03$\frac{J}{gC}$ × 49.3 g× 25 °C

Solving:

Q1=2501.975 J=2.501975 kJ≅ 2.50 kJ

• Change of state

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.

In this case, you know:

• m= 49.3 g
• ΔHfus= ?  

Replacing:

Q2= 49.3 g× ΔHfus

• 0 °C to 44.7 °C

Similar to sensible heat previously calculated, you know:

• c(liquid)= 4.18 $\frac{J}{gC}$  
• m= 49.3 g
• ΔT= Tfinal - Tinitial= 44.7°C - 0°C= 44.7 °C

Replacing:

Q3= 4.18$\frac{J}{gC}$ × 49.3 g× 44.7 °C

Solving:

Q3= 9211.5078 J=9.2115078 kJ≅ 9.21 kJ

• Total heat required

The total heat required is calculated as:

Total heat required= Q1 + Q2 + Q3

Total heat required= 2.50 kJ + 49.3 g× ΔHfus + 9.21 kJ

Heating a 49.3 g sample of g ice from -25.0 °C to water at 44.7 °C requires 28.23 kJ of heat. This is, the total heat required is 28.23 kJ. Then:

28.23 kJ= 2.50 kJ + 49.3 g× ΔHfus + 9.21 kJ

Solving:

28.23 kJ= 11.71 kJ + 49.3 g× ΔHfus

28.23 kJ- 11.71 kJ = 49.3 g× ΔHfus

16.52 kJ = 49.3 g× ΔHfus

16.52 kJ ÷ 49.3 g= ΔHfus

0.335091$\frac{kJ}{g}$= 335.091 $\frac{J}{g}$ =ΔHfus

In summary, the heat of fusion of water is 335.091 $\frac{J}{g}$ .

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