Question

Work done in converting 1g of ice at $-10{}^\circ C$ into steam at $100{}^\circ C$ is
A. 3.04KJ
B. 6.05KJ
C. 0.721KJ
D. 0.616KJ

Answer 1

This question can be dealt in four different steps. Firstly you could find the heat absorbed in increasing the temperature of ice from $-10{}^\circ C$ to$0{}^\circ C$. Then you could find the heat supplied in changing the state from ice to water. After that you could find the heat used for increasing the temperature to $100{}^\circ C$and finally, you could find the heat required for conversion into steam. The sum of all these heat supplies gives you the answer.
Formula used:
Heat absorbed,
$Q=mc\Delta T$
$Q=m{{L}_{f}}$
$Q=m{{L}_{v}}$

Complete answer:
In the question we are given 1g of ice at $-10{}^\circ C$ and we are supposed to find the work done to convert this into steam at$100{}^\circ C$.
We have to discuss this question in three steps. We have to first increase the temperature of ice from $-10{}^\circ C$ to$0{}^\circ C$. The heat absorbed in this process can be found by using,
${{Q}_{1}}=mc\Delta T$
Where, m is the mass of the ice, c is its specific heat capacity and $\Delta T$ is the change in temperature. Substituting the value we get,
${{Q}_{1}}=1\times 2.1\times \left( 0-\left( -10 \right) \right)=2.1\times 10$
$\therefore {{Q}_{1}}=21J$ …………………………………. (1)
Now we have to find the heat supplied to change the state of ice to water and this can be given by,
${{Q}_{2}}=m{{L}_{f}}$
Substituting the values we get,
${{Q}_{2}}=1\times 339=339J$ ………………………………….. (2)
Heat absorbed by the water so as increase its temperature from $0{}^\circ C$ to$100{}^\circ C$,
${{Q}_{3}}=1\times 4.18\times \left( 100-0 \right)=418J$……………………………………. (3)
Now, the heat absorbed for changing the state of water to steam can be given by,
${{Q}_{4}}=m{{L}_{v}}$
${{Q}_{4}}=1\times 2260=2260J$ ………………………………………… (4)
The net heat absorbed for the whole conversion could be given by,
$Q={{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}+{{Q}_{4}}$
$\Rightarrow Q=21+339+418+2260$
$\therefore Q=3038J=3.04kJ$
Therefore, we found the net work done in converting 1g of ice at $-10{}^\circ C$ into steam at $100{}^\circ C$ to be 3.04kJ.

Hence, option A is found to be the answer.

Note:
Chances are there that you may miss out certain steps while finding the answer. So, make sure that you always follow the same order of steps while dealing with questions like this. Otherwise, you will end up with a different answer. Also, note that the temperatures to which we have raised, that is, $0{}^\circ C$ and $100{}^\circ C$ are the melting point of ice and the boiling point of water.