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Questions on calorimetry without phase change

Question

A copper calorimeter of mass 1.5 kg has 200 g of water at $25^{0} C$ . How much water (approx.) at $50^{0} C$ is required to be poured in the calorimeter, so that the equilibrium temperature attained is $40^{0} C ?$ (Sp. heat capacity of copper = 390 J/kg- $^{0} C$ )

Moderate

A

508.3 gram
B

450 gram
C

370 gram
D

610 gram

Solution

Let ‘m’ be the required mass of water.
Then heat lost by hot water
$Q_{lost} = (m) (1 \frac{cal}{g -^{0} C}) {(50 - 40)}^{0} C - - - - - - (i)$
Heat gained by the calorimeter
$Q_{1} = [(1500 g) (\frac{390}{4.2 \times 10^{3}} \frac{cal}{g -^{0} C)})] {(40 - 25)}^{0} C$
Heat gained by the water present in it
$Q_{2} = [(200 g) (1 \frac{cal}{g -^{0} C})] {(40 - 25)}^{0} C$
Total heat gained by the calorimeter and the water present in it
$Q_{gain} = Q_{1} + Q_{2} - - - - - - - (ii)$
From principle of calorimetry
heat lost by hot body = heat gained by cold body
So, equating Eqs. (i) and (ii) and solving for m, we get
m = 508.93 gram

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