Question

200 g of hot water at $80^\circ C$ is added to $300g$ of cold water at $10^\circ C$. Neglecting the heat taken by the container, calculate the final temperature of the mixture of the water.
Specific heat capacity of water= $4200JK{g^{ - 1}}{K^{ - 1}}$.
A.) $48^\circ C$
B.) $18^\circ C$
C.) $28^\circ C$
D.) $38^\circ C$

Answer 1

Hint- This is a question from the topic transfer of heat. According to the principle of transfer of heat for a mixture, the amount of heat that is lost is always equal to the amount of the heat gained during the formation of the particular mixture.

Step By Step Answer:
Given quantities in the question,

Mass of the hot water=$${m_h} =$$$200g = 0.2g$Temperature of the hot water =${T_h} = 80^\circ C$Mass of the cold water =$300g = 0.3g$Temperature of the cold water=${T_c} = 10^\circ C$Specific heat capacity of water=$4200JK{g^{ - 1}}{K^{ - 1}}$.

And let us suppose that the final temperature of the mixture= $T^\circ C$

So, from the above given quantities we can find the heat lost and heat gained during the formation of the mixture.

So, heat lost by the hot water = ${Q_h} = mC\Delta T$
${Q_h} = 0.2 \times 4200 \times \left( {80 - T} \right)$ -----equation (1)

And similarly, with the help of the same formula,

The heat gained by the cold water = ${Q_c} = 0.3 \times 4200 \times \left( {T - 10} \right)$------equation (2)

Now as we know, according to the principle of mixtures,

Heat lost by the hot water= heat gained by the cold water
So now equating the equation (1) and equation (2), we get
$0.2 \times 4200 \times \left( {80 - T} \right) = 0.3 \times 4200 \times \left( {T - 10} \right)$
$\Rightarrow 16 - 0.2T = 0.3T - 3$
$\Rightarrow 16 + 3 = 0.3T + 0.2T$
$\Rightarrow 0.3T + 0.2T = 16 + 3$
$\Rightarrow 0.5T = 19$
$\Rightarrow T = \dfrac{{19}}{{0.5}} = 38$
So finally, we have got $T = 38^\circ C$.

Hence the final temperature of the mixture of the water is $38^\circ C$.

So, option (D) is the correct answer to the given question.

Note- There is another method to find the answer directly with a formula which is written as follows,
The final temperature of the mixture, $T = \dfrac{{{m_h}{T_h} + {m_c}{T_c}}}{{{m_h} + {m_c}}}$,
Where the notations are as mentioned above in the solution.
Now if we go with this formula then we will find the solution with this formula in simple calculations. So, we will go as follows,
$T = \dfrac{{(200 \times 80) + (300 \times 10)}}{{(200 + 300)}} = \dfrac{{19000}}{{500}} = 38$
So, we have $T = 38^\circ C$.