Heat And Thermodynamics Ques 33
- A metal ball of mass $0.1 kg$ is heated upto $500^{\circ} C$ and dropped into a vessel of heat capacity $800 JK^{-1}$ and containing 0.5 $kg$ water. The initial temperature of water and vessel is $30^{\circ} C$. What is the approximate percentage increment in the temperature of the water? [Take, specific heat capacities of water and metal are respectively $4200 Jkg^{-1} K^{-1}$ and $400 Jkg^{-1} K^{-1}$ ]
(2019 Main, 11 Jan II)
(a) $25 \text{\%}$
(b) $15 \text{\%}$
(c) $30 \text{\%}$
(d) $20 \text{\%}$
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Answer:
Correct Answer: 33.(d)
Solution:
Formula:
- Using heat lost or gained without change in state is $\Delta Q=m s \Delta T$, where $s$ is specific heat capacity and $T=$ change in temperature
Let final temperature of ball be $T$.
Then heat lost by ball is,
$$ \Delta Q=0.1 \times 400(500-T) $$
This lost heat by ball is gained by water and vessel and given as
Heat gained by water,
$$ \Delta Q _1=0.5 \times 4200(T-30) $$
and heat gained by vessel is
$$ \begin{aligned} \Delta Q _2 & =\text { heat capacity } \times \Delta T \\ & =800 \times(T-30) \end{aligned} $$
According to principle of calorimetry, total heat lost $=$ total heat gained
$$ \begin{aligned} \Rightarrow & 0.1 \times 400(500-T) \\ = & 0.5 \times 4200(T-30)+800(T-30) \end{aligned} $$
$$ \begin{array}{ll} \Rightarrow & (500-T)=\frac{(2100+800)(T-30)}{40} \\ \Rightarrow & 500-T=72.5(T-30) \\ \Rightarrow & 500+217.5=72.5 T \text { or } T=36.39 K \end{array} $$
So, percentage increment in temperature of water
$$ \text{\%}=\frac{36.39-30}{30} \times 100 \approx 20 \text{\%} $$
BETA
