Question

200 g of hot water at ${{80}^{0}}C$ is added to $300g$ of cold water at ${{10}^{0}}C$. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water. Specific heat capacity of water$=4200Jk{{g}^{-1}}{{K}^{-1}}$.

Answer 1

Hint: When cold and hot water are mixed together, the cold water absorbs heat from the hot water, thereby increasing its temperature while there is a drop in the temperature of the hot water. Finally, at equilibrium the whole mixture comes to a single final temperature. This temperature is decided by the respective initial masses of hot and cold water and the specific heat capacity of water.

Formula used: When cold water of mass ${{m}_{C}}$ and temperature ${{T}_{C}}$ is mixed with hot water of mass ${{m}_{H}}$ and temperature ${{T}_{H}}$, heat is lost by the hot water and gained by the cold water. Thus, the mixture comes to a single final temperature $T$ at equilibrium.

$\text{Heat lost by hot water = }{{\text{m}}_{H}}S\left( {{T}_{H}}-T \right)$
$\text{Heat gained by cold water = }{{\text{m}}_{C}}S\left( T-{{T}_{C}} \right)$

Where $S$ is the specific heat capacity of water, that is, the amount of heat required to raise the temperature of $1kg$ of water by $1K$.

Complete step by step answer:

When hot water and cold water are mixed together, the hot water starts losing heat and its temperature starts dropping. This lost heat is gained by the cold water and its temperature starts increasing. Finally, the whole mixture comes to a single temperature lying between the initial temperatures of the hot and cold water respectively.

Now,

When cold water of mass ${{m}_{C}}$ and temperature ${{T}_{C}}$ is mixed with hot water of mass ${{m}_{H}}$ and temperature ${{T}_{H}}$, heat is lost by the hot water and gained by the cold water. Thus, the mixture comes to a single final temperature $T$ at equilibrium.
$\text{Heat lost by hot water = }{{\text{m}}_{H}}S\left( {{T}_{H}}-T \right)$--(1)
$\text{Heat gained by cold water = }{{\text{m}}_{C}}S\left( T-{{T}_{C}} \right)$ --(2)

Where $S$ is the specific heat capacity of water, that is, the amount of heat required to raise the temperature of $1kg$ of water by $1K$.

Since, the heat lost by the hot water is assumed to be completely absorbed by the cold water in the question (since heat taken by the container is neglected). Therefore,
$\text{Heat gained by cold water = Heat lost by hot water}$

Hence, using (1) and (2), we get,

$\therefore {{m}_{C}}S\left( T-{{T}_{C}} \right)={{m}_{H}}S\left( {{T}_{H}}-T \right)$
$\therefore {{m}_{C}}\left( T-{{T}_{C}} \right)={{m}_{H}}\left( {{T}_{H}}-T \right)$ --(3)

Now, let us analyse the question.

Mass of cold water ${{m}_{C}}=300g$

Initial temperature of cold water ${{T}_{C}}={{10}^{0}}C=283K$ $\left( \because K=273{{+}^{0}}C \right)$where $K$ implies Kelvin temperature.

Mass of hot water ${{m}_{H}}=200g$

Initial temperature of hot water ${{T}_{H}}={{80}^{0}}C=353K$ $\left( \because K=273{{+}^{0}}C \right)$ where $K$ implies Kelvin temperature.

Specific heat of water$S=4200Jk{{g}^{-1}}{{K}^{-1}}$.
We have to find the final temperature of the mixture$T$.
Plugging these values in equation (3), we get,

$300\left( T-283 \right)=200\left( 353-T \right)$
$\Rightarrow 3\left( T-283 \right)=2\left( 353-T \right)$
$\Rightarrow 3T-849=706-2T$
$\Rightarrow 3T+2T=706+849$
$\Rightarrow 5T=1555$
$\Rightarrow T=\dfrac{1555}{5}=311K={{\left( 311-273 \right)}^{0}}C={{38}^{0}}C$

Hence, the final temperature of the mixture is ${{38}^{0}}C$.

Note: Students often forget to convert the temperature into the Kelvin scale and proceed with the Celsius or Fahrenheit temperature. This mistake could also have been made by a student in this problem. This leads to a completely different and wrong answer since. In thermodynamics all problems must be solved in the SI unit that is Kelvin, else the answer arrived at will be wrong.Therefore, it is always a good practice to first convert all temperatures and other quantities to the SI units and then proceed with the problem and its calculation part. If the question requires an answer in some other unit, first an answer should be arrived at in the Kelvin scale and then converted to the required scale, as done in the above problem. Students might be thinking that there is no requirement of the specific heat capacity given in the question, since it gets cancelled anyway. However, if the liquids that were mixed were not the same, then each liquid would have a different specific heat capacity. Then they would not get cancelled and play a role in the calculation process.