2,kg of metal at 100,^oC is cooled by 1,kg of water at 0,^oC . If specific heat capacity of metal is

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10-1.Thermometry, Thermal Expansion and Calorimetry

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$2\,kg$ of metal at $100\,^oC$ is cooled by $1\,kg$ of water at $0\,^oC$ . If specific heat capacity of metal is $\frac {1}{2}$ of specific heat capacity of water, final temperature of mixture would be

A

$50\,^oC$

B

More than $50\,^oC$

C

Less than $50\,^oC$

D

None of the above

Solution

heat loss by metal $=$ heat gain by water $\mathrm{m}_{1} \mathrm{s}_{1}(100-\theta)=\mathrm{m}_{2} \mathrm{s}_{2}(\theta-0)$

$\Rightarrow 2 \times \frac{s_{2}}{2} \times(100-\theta)=1 \times s_{2} \times \theta$

$\Rightarrow 100-\theta=\theta \quad \Rightarrow \theta=50^{\circ} \mathrm{C}$

Standard 11

Physics

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