Heat and Thermodynamics 1 Question 7
7. An unknown metal of mass $192 g$ heated to a temperature of $100^{\circ} C$ was immersed into a brass calorimeter of mass $128 g$ containing $240 g$ of water at a temperature of $8.4^{\circ} C$. Calculate the specific heat of the unknown metal, if water temperature stabilises at $21.5^{\circ} C$. (Take, specific heat of brass is $394 J kg^{-1} K^{-1}$ )
(2019 Main, 10 Jan II)
(a) $916 J kg^{-1} K^{-1}$
(b) $654 J kg^{-1} K^{-1}$
(c) $1232 J kg^{-1} K^{-1}$
(d) $458 J kg^{-1} K^{-1}$
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Answer:
Correct Answer: 7. (a)
Solution:
- Key Idea
The principle of calorimetry states that total heat lost by the hotter body equals to the total heat gained by colder body, provided that there is no exchange of heat with the surroundings.
Let specific heat of unknown metal is $s$ and heat lost by this metal is $\Delta Q$.
Heat lost and specific heat of a certain material/substance are related as
$$ \Delta Q=m s \Delta T \cdots(i) $$
For unknown metal, $m=192 g$ and
$$ \begin{aligned} \Delta T & =(100-21.5)^{\circ} C \\ \therefore \quad \Delta Q^{\prime} & =192(100-21.5) \times s \quad \cdots (ii) \end{aligned} $$
Now, this heat is gained by the calorimeter and water inside it.
As, heat gained by calorimeter can be calculated by Eq. (i). So, for brass specific heat,
$$ \begin{aligned} s & =394 J kg^{-1} K^{-1} \\ & =0.394 J g^{-1} K^{-1} \end{aligned} $$
Mass of calorimeter, $m=128 g$
Change in temperature, $\Delta T=(21.5-8.4)^{\circ} C$
So, using Eq. (i) for calorimeter, heat gained by brass
$$ \Delta Q _1=128 \times 0.394 \times(21.5-8.4) $$
Heat gained by water can be calculated as follows mass of water, $m=240 g$,
specific heat of water, $s=4.18 J g^{-1} K^{-1}$, change in temperature, $\Delta T=(21.5-8.4)^{\circ} C$
Using Eq. (i) for water also, we get
heat gained by water,
$$ \Delta Q _2=240 \times 4.18 \times(21.5-8.4) $$
Now, according to the principle of calorimeter, the total heat gained by the calorimeter and water must be equal to heat lost by unknown metal
$$ \Delta Q^{\prime}=\Delta Q _1+\Delta Q _2 $$
Using Eqs. (ii), (iii) and (iv), we get
$$ \begin{array}{rlrl} \Rightarrow & & =192(100-21.5) \times s \\ & =128 \times 0.394 \times(21.5-8.4)+240 \\ \Rightarrow & & \quad 15072 s & =660.65+13142 \\ \Rightarrow & & s & =0.916 J g^{-1} K^{-1} \\ & & \\ \text { or } & & s & =916 J Kg^{-1} K^{-1} . \end{array} $$
