Question

How much heat energy (in megajoules) is needed to convert 7 kilograms of ice at –9°C to water at 0°C?

Answer 1

1.       Energy released from changing the phase of a substance from the liquid phase to solid phase can be calculated by using the specific latent heat of fusion. The heat of fusion of water at 0 degrees Celsius is 334 J/g. Calculation are as follows:

 

 Energy = 7000 grams x 334 J/g

 Energy = 2338000 J or 2.38 MJ

Answer 2

Answer;

A. 2.48 MJ

Explanation;

-Heat energy may be given by the product of specific heat capacity , mass and change in temperature, or the product of heat capacity and change in temperature.

-When a substance is changing from solid state to liquid state then the heat energy is calculate by multiplying the latent heat of fusion of a substance by the mass. Latent heat of fusion of ice =3.34×10⁵J/kg and specific heat capacity of water is 2100 J/kg/K.

Specific heat energy: Q = mcΔT

Latent heat energy: Q = mLf

In this case the heat energy required will be;

Heat required to melt ice and raise its temperature to 0°C:

Q₁ = mLf+mciΔT

Q₁ = (7 kg*3.34×10⁵J/kg) + ((7kg*2100J/kgK*{0°C-(9°C)})

= 2.48 MJ (MEGA JOULES)