Question

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When 100 g of boiling water at 100°C is added into a calorimeter containing 300g of cold water at 10°C, temperature of the mixture becomes 20°C. Then a metallic block of mass 1 kg at 10°C is dipped into the mixture in the calorimeter. After reaching thermal equilibrium, the final temperature becomes 19°C. What is the specific heat of the metal?
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Answer 1

Explanation:

As per the given data,

100g of water at 100 °C is added into a calorimeter containing 300g of cold water at 10°C so the temperature of the combined mixture becomes 20°C

• Mass of the mixture(m) = 400g

Now a metallic block of mass 1 kg(M) at 10 °C is dipped into the above mixture.

The final temperature of the combination (block + mixture) is found to be 19°C.

From the above statement, we can easily conclude that the heat gained by the metal block is obtained from the heat lost by the mixture thus gaining thermal equilibrium.

$\to\sf{Heat \: gain= Heat\: loss}$

$\to\sf{M \times S \times \Delta T = m \times s \times \Delta T}$

$\to\sf{1000 \times S \times (19-10) = 400 \times 1 \times (20-19)}$

$\to\sf{1000 \times S \times 9 = 400 \times 1 \times 1}$

$\to\sf{S = \dfrac{4}{90}}\\ \\ \to\sf{S = 0.044 \: Cal ^{\circ}{C}^{-1} g^{-1}}$

The specific heat of the metal is 0.044 Cal C⁻¹ g⁻¹.