Question

A vessel contains 110g of water. The heat capacity of the vessel is equal to 10g of water. The initial temperature of water in the vessel is $10{}^\circ C$. If 220g of hot water at $70{}^\circ C$ is poured in the vessel, the final temperature (neglecting radiation loss) will be
A. $70{}^\circ C$
B. $80{}^\circ C$
C. $60{}^\circ C$
D. $50{}^\circ C$

Answer 1

As a first step, we could note down all the given values from the question. Then you could recall the expression for heat transferred. After that you could recall how the law of conservation of energy affects the heat transfer. Then you could make the substitutions accordingly and thus find the answer.
Formula used:
Heat transferred,
$Q=mc\Delta T$

Complete answer:
In the question, we are given 110g of water and the heat capacity of the vessel is equal to that of 10g of water and the initial temperature of water in a vessel to be$10{}^\circ C$. We are supposed to find the final temperature of the water when 220g of hot water at $70{}^\circ C$ is poured into the vessel.
Firstly, let us assume that T is the resultant temperature of the mixture. We know that heat transferred for a substance of mass m and specific heat capacity c to raise the temperature by $\Delta T$ is given by,
$Q=mc\Delta T$
By law of conservation of energy we know that the heat lost by water of $70{}^\circ C$ is equal to the heat gained by water at$10{}^\circ C$.
${{m}_{1}}c\left( {{T}_{1}}-T \right)={{m}_{2}}c\left( T-{{T}_{2}} \right)$
Where, ${{m}_{1}}=220g$
${{T}_{1}}=70{}^\circ C$
${{T}_{2}}=10{}^\circ C$
${{m}_{2}}=110+10=120g$
Substituting these values we get,
$220\times 1\times \left( 70-T \right)=120\times 1\times \left( T-10 \right)$
$\Rightarrow 22\left( 70-T \right)=12\left( T-10 \right)$
$\Rightarrow 1660=34T$
$\therefore T=48.82{}^\circ C$
Therefore, we found the resultant temperature of the mixture to be,
$T=48.82{}^\circ C\approx 50{}^\circ C$

So, the correct answer is “Option D”.

Note: In the question, we are given that the specific heat capacity of the vessel is equal to that of 10g of water. So, we have accordingly performed the substitutions. Also, we know that the specific heat capacity of water is given by,
$c=1cal/g{}^\circ C$