Question

What mass of steam at ${{100}^{\circ }}C$ must be mixed with 150g of ice at melting point in a thermally insulated container to produce liquid water is:
A: 40g
B: 150g
C: 30g
D: 100g

Answer 1

To solve this problem, we have to calculate the energies of the elements that are relative to ice that is ${{0}^{\circ }}C$of water. Since the container is thermally insulated, there is no heat loss. Hence, the heat needed to melt ice and raise water temperature is the same as the heat needed to transfer steam to water and cool it.

Formula used:
$Q=mL$ where Q is the heat energy absorbed per mass m and L is the latent heat of the material.
$Q=mc\Delta T$ where Q is the heat energy absorbed per mass m, c is the specific heat of the material and $\Delta T$ is the change in material.

Complete step by step solution:
In the question, we are given that mass of steam at ${{100}^{\circ }}C$ is to be mixed with 150g of ice at melting point in a thermally insulated container. From the given conditions we can infer that the heat energy absorbed per mass by both ice and steam are the same that is,
${{Q}_{ice}}={{Q}_{steam}}$
${{Q}_{ice}}={{L}_{f}}{{m}_{i}}+{{m}_{i}}{{c}_{w}}\Delta T \\\ =0.15kg\times .333\dfrac{KJ}{kg}+0.15kg\times 4.187\dfrac{KJ}{kg.K}\times 60=87.6KJ \\\$
${{Q}_{steam}}={{L}_{v}}.{{m}_{s}}+{{m}_{s}}{{c}_{w}}\Delta T$
We know that ${{Q}_{ice}}={{Q}_{steam}}$
Combining the equations, we get
{{Q}_{ice}}={{L}_{v}}.{{m}_{s}}+{{m}_{s}}{{c}_{w}}\Delta T \\\ ={{m}_{s}}({{L}_{v}}+{{c}_{w}}\Delta T) \\\
$\therefore{{m}_{s}}=\dfrac{{{Q}_{ice}}}{({{L}_{v}}+{{c}_{w}} T)}$=$\dfrac{87.6kJ}{2256\dfrac{KJ}{kg}+4.187\dfrac{KJ}{kg.K}\times 60}=30g \\\$

Hence, the required mass of steam is 30g.

So, the correct answer is “Option C”.

Note:
Latent heat of ice is termed as latent heat of fusion and latent heat of steam is termed as latent heat of vaporisation.
Also, the standard values of the specific heat and the latent heat is being used in our question.
we can infer that the heat energy absorbed per mass by both ice and steam are the same.