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

Using the equation , q = m.s.t
                        q = 7 x 4.2 x 1000 x 9 joules
                              264600 joules = 0.2646 mega joules

Answer 2

Answer:

The heat energy is 2.48 mega joules.

Explanation:

To find : How much heat energy (in mega joules) is needed to convert 7 kilograms of ice at -9°C to water at 0°C?

Solution :

The heat energy in this case formula is given by,

$Q=Q_1+Q_2$

$Q= mL_f+mc_i\Delta{T}$

Where,

$Q_1= mL_f$ is the Latent heat energy.

$Q_2=mc_i\Delta{T}$ is the Specific heat energy.

We have given,

Mass m=7 kilograms

Latent heat of fusion of ice is $L_f=3.34\times10^5\ J/kg$

Specific heat capacity of water is $c_i=2100 m\ J/kg/K$

$\Delta T$ is the change in the temperature,

Substitute the value in the formula,

$Q=7\times 3.34\times10^5+7\times 2100\times (0^\circ C-(-9^\circ C))$

$Q=7\times 3.34\times10^5+7\times 2100\times 9$

$Q=2338000+132300$

$Q=2470300$

or $Q=2.48MJ$

Therefore, The heat energy is 2.48 mega joules.