Calculating Specific Latent Heat (SQA National 5 Physics): Revision Note

Water ice melts and freezes at its melting point.

Specific latent heat of fusion of water = $3.34\times {10}^5\mathrm{J}{\mathrm{kg}}^{-1}$

Specific latent heat of vaporisation of $22.6\times {10}^5\mathrm{J}{\mathrm{kg}}^{-1}$

(i) Calculate the minimum amount of energy required to melt 1.3 kg of water ice at its melting point.

(ii) State the amount of heat energy given out as 1.3 kg of water freezes into ice at its melting point.

Answer (i):

Step 1: Write down the relevant relationship

$E_h=ml$

Step 2: Substitute in the known values to calculate

• Melting ice is changing state from a solid to a liquid

• Therefore, the value for specific latent heat of fusion is used

$E_h=1.3\times \left(3.34\times {10}^5\right)$

$E_h=4.342\times {10}^5\mathrm{J}$

Step 3: Round to an appropriate number of significant figures

• The least precise data point is 2 s.f. (1.3 kg)

• Therefore, the answer must be given to the same precision

$E_h=4.3\times {10}^5\mathrm{J}\left(2\mathrm{s}.\mathrm{f}.\right)$

Answer (ii):

• If $4.3\times {10}^5\mathrm{J}$ of heat energy is required to melt 1.3 kg of ice into liquid water, then the same amount of heat energy is given out when 1.3 kg of liquid water freezes into ice at its melting point

• Therefore, $4.3\times {10}^5\mathrm{J}$ is given out

Both the relationship and the latent heat values will be provided to you in the exam. You do not need to remember any of the specific values.