Question

400 g of vegetable oil of specific heat capacity \[1.98\,{J}/{g{}^\circ C}\;\]is cooled from \[100{}^\circ C\]. Find the final temperature, if the heat energy given out by oil is 47376 J.
A. \[30.2{}^\circ C\]
B. \[40.2{}^\circ C\]
C. \[50.2{}^\circ C\]
D. \[43.2{}^\circ C\]

Answer 1

Hint: Using the formula of specific heat capacity of a substance, this problem can be solved. All the other parameters, except the final temperature is given, so substitute the given values in the formula and find the required value of the final temperature.

Formula used: \[Q=mc\Delta T\]

Complete step by step answer:
From given, we have the data,
The specific heat capacity of the vegetable oil, \[c=1.98{J}/{g{}^\circ C}\;\]
The mass of the vegetable oil, m = 400 g
The heat energy given out by the oil, Q = 47376 J
The initial temperature of the oil, \[{{T}_{1}}=100{}^\circ C\]
The specific heat capacity of a substance is the heat required to raise the temperature of the unit mass of that substance by one degree (usually).
The formula for calculating the value of the specific heat capacity of a substance is given as follows.
\[Q=mc\Delta T\]
Where Q is the heat energy, m is the mass of the substance, c is the specific heat capacity and \[\Delta T\]is the change in the temperature.
As the given parameters have the same units, so there is no need of unit conversion of the parameters. Thus, continue with the further calculation.
Substitute the given values in the above equation, to find the value of the final temperature.
So, we have,
\[\begin{align}
  & 47376=400\times 1.98\times (100-{{T}_{2}}) \\
 & \Rightarrow 100 - {{T}_{2}}=\dfrac{47376}{792} \\
\end{align}\]
Continue the further calculation to obtain the value of the final temperature.
\[\begin{align}
  & {{T}_{2}}=100-\dfrac{47376}{792} \\
 & \Rightarrow {{T}_{2}}=40.2{}^\circ C \\
\end{align}\]
As the value of the final temperature obtained is \[40.2{}^\circ C\]

So, the correct answer is “Option B”.

Note: The things to be on your finger-tips for further information on solving these types of problems are: The units of the given parameters should be taken care of, as, the given units of the parameters are in terms of joule, so no need of conversion, otherwise, if given in terms of calorie, get it converted to joule.