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A baker contains \[200{{ }}gm\] of water. The heat capacity of the baker is equal to that of \[20{{ }}gm\] of water. The initial temperature of the water in the baker is $20^\circ C$. If \[440gm\] of hot water at $92^\circ C$ is poured in it, the final temperature, neglecting radiation loss, will be nearest to:
A) $58^\circ C$
B) $68^\circ C$
C) $73^\circ C$
D) $78^\circ C$

Last updated date: 29th May 2024
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Hint: The heat dissipated from a hot body is equal to heat absorbed by the cold body. Here, the water is the hot body that will dissipate heat energy. this heat energy will be absorbed by the baker and then the entire water-baker system will come to a point where there will be no heat transfer.

Complete step by step solution:
Let’s define all the terms given in the question
The initial content of the water in the baker = \[200{{ }}gm\]
The initial temperature of the water in the baker = $20^\circ C$
Amount of poured water to the baker = \[440gm\]
The temperature of the poured water to the baker = $92^\circ C$
From the principle of calorimetry, we know the total heat gained by a cold body will be equal to the heat lost by the hot body.
That is, Heat lost = Heat gained
For the calculation purposes, we use 1 to indicate the poured water, 2 for the initial water content in the baker, 3 for the baker.
We know, Heat lost = Heat gained
That is,
\[{m_1}{s_1}\Delta {T_1} = {m_2}{s_2}\Delta {T_2} + W\Delta {T_3}\]………………… (1)
Where, $m$is the mass of the content
$s$ is the specific heat capacity of the content
$\Delta T$is the change in temperature of the content
$W$is the water equivalent of the baker (Given in the question as $20$ times)
Let’s take the final temperature as $T$
$\therefore \Delta {T_1} = 92 - T$
$ \to \Delta {T_2} = T - 20$
$ \to \Delta {T_3} = T - 20$
Applying these values to the equation (1), we get,
$440 \times 1 \times (92 - T) = 200 \times 1 \times (T - 20) + 20 \times (T - 20)$
$ \Rightarrow 440 \times (92 - T) = 220 \times (T - 20)$
$ \Rightarrow 2 \times (92 - T) = T - 20$
$ \Rightarrow 184 - 2T = T - 20$
$ \Rightarrow 204 = 3T$
$ \Rightarrow T = 68$
That is the final temperature is $68^\circ C$

So, the final answer is option (B), $68^\circ C$.

Note: The specific heat capacity is the amount of energy that must be added, in the form of heat, to one unit of mass of a substance in order to cause an increase of one unit in temperature. The specific heat of water is given by, \[1 cal / °C-kg = 4.184 J/K-kg\].