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Standard VIII

Physics

Specific Heat

A piece of ir...

Question

A piece of iron weighing $100$ g was heated to ${98.5}^{o}$ C and dropped in a calorimeter weighing $46$ g and containing $85.4$ g water at $15^{o}$ C. The temperature was found to be $22^{o}$ C. Calculate the specific heat of iron if the specific heat of material of calorimeter is $0.1$ .

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Solution

Given: A piece of iron weighing $100 g$ was heated to ${98.5}^{o} C$ and dropped in a calorimeter weighing $46 g$ and containing $85.4 g$ water at $15^{o} C$ . The temperature was found to be $22^{o} C$ .
To find the specific heat of iron if the specific heat of material of calorimeter is 0.1.
Solution:
As per the given criteria,
Mass of iron, $m_{i} = 100 g$
Mass of the water in calorimeter, $m_{w} = 85.4 g$
Water equivalent of the calorimeter, $w_{e} = 46 \times 0.1 = 4.6 g$
Temperature of iron, $T_{i} = {98.5}^{\circ} C$
Temperature of water or calorimeter, $T_{w} = 15^{\circ} C$
Final temperature, $T = 22^{\circ} C$
Heat lost by the iron = mass\times specific heat\times fall in temperature
$H_{L} = m s Δ T ⟹ H_{L} = 100 \times s \times (98.5 - 22) = 7650 s$
Heat gained by the calorimeter,
$H_{g} = m s Δ T ⟹ H_{g} = 4.6 \times 1 \times (22 - 15) ⟹ H_{g} = 32.2 c a l$
Heat gained by water,
$H_{w} = m s Δ T ⟹ H_{w} = 85.4 \times 1 \times (22 - 15) ⟹ H - w = 597.8 c a l$
Therefore total heat gained = 32.2+597.8=630cal
According to the principle of calorimetery,
Heat lost=Heat gained
$7650 s = 630 ⟹ s = \frac{630}{7650} ⟹ s = 0.0823$
is the required specific heat of iron.

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A piece of iron weighing 100g was heated to 98.5oC and dropped in a calorimeter weighing 46g and containing 85.4g water at 15oC. The temperature was found to be 22oC. Calculate the specific heat of iron if the specific heat of material of calorimeter is 0.1.

A piece of iron weighing $100$ g was heated to ${98.5}^{o}$ C and dropped in a calorimeter weighing $46$ g and containing $85.4$ g water at $15^{o}$ C. The temperature was found to be $22^{o}$ C. Calculate the specific heat of iron if the specific heat of material of calorimeter is $0.1$ .