Question

A meat ball of mass $0.1kg$ is heated up to ${500^ \circ }C$ and dropped into a vessel of heat capacity $800J{K^{ - 1}}$ and containing $0.5kg$ water. Find the approximate increment in the temperature of water if the initial temperature of water and vessel is ${30^ \circ }C$. It is given that specific heat capacities of water and metal are $4200Jk{g^{ - 1}}{K^{ - 1}}$ and $400Jk{g^{ - 1}}{K^{ - 1}}$.
A. $30\% $
B. $20\% $
C. $25\% $
D. $15\% $

Answer 1

Hint: Use the principle of calorimetry to find the final value of temperature. Then, calculate change in temperature from its initial value to final value and calculate its percentage increment by dividing change in temperature by initial temperature.

Complete step by step answer:
Calorimetry can be defined as the act of calculating the changes in the state variables of the body so that we can calculate the heat transfer associated with changes of its states under specific conditions. This is performed by using Calorimeter.
In this question, we have to use the principle of calorimetry to find the final temperature of water. So, the principle of calorimeter states that, “in an isolated system, the heat energy gained by the cold body is equal to the heat energy lost by the hot body”. It can represented as –
$ {m_1}{c_1}\left( {{t_1} - t} \right) = {m_2}{c_2}\left( {t - {t_2}} \right)$
From the question, using the principle of calorimetry and putting the values in the above equation, we get –
$0.1 \times 400 \times \left( {500 - t} \right) = 0.5 \times 4200 \times \left( {t - 30} \right) + 800\left( {t - 30} \right)$
Now, solving the above equation –
$
   \Rightarrow 40\left( {500 - t} \right) = 2100\left( {t - 30} \right) + 800\left( {t - 30} \right) \\
   \Rightarrow 40\left( {500 - t} \right) = \left( {2100 + 800} \right)\left( {t - 30} \right) \\
   \Rightarrow 20000 - 40t = 2900t - 87000 \\
   \Rightarrow 2940t = 107000 \\
  \therefore t = {36.4^ \circ }C \\
 $
Now, calculating the change in temperature of water –
$
  \therefore \Delta t = t - {30^ \circ }C \\
   \Rightarrow \Delta t = 36.4 - 30 \\
   \Rightarrow \Delta t = {6.4^ \circ }C \\
 $
Calculating the approximate increment in the temperature of water –
$
   \Rightarrow \dfrac{{\Delta t}}{t} = \dfrac{{6.4}}{{30}} \\
  \therefore \dfrac{{\Delta t}}{t} \approx 0.2 \\
 $
To calculate the percentage of increment in water we have to multiply the $\dfrac{{\Delta t}}{t}$ with $100$ -
$
   \Rightarrow \dfrac{{\Delta t}}{t} \times 100 \\
   \Rightarrow 0.2 \times 100 \\
   \Rightarrow 20\% \\
 $
Hence, the approximate increment in the temperature of water is $20\% $.

Therefore, the correct option is (B).

Note:Be careful when using the principle of calorimeter. As the metal ball is dropped in the vessel which contains water, the heat gained by the ball is equal to the sum of heat lost by vessel and heat lost by the water.
The heat transfer occurs when bodies have the same temperature.