A block of metal of mass 0-20 kg is heated to 150-C and dropped in copper calorimeter of water equivalent 0-025 kg- containing 150 g of water at 27-C- The final temperature is 40-C- Find the specific heat of the metal-Given- specific heat of water - 4-2- 103J-kg-1 K-1-

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Byju's Answer

Standard IX

Physics

Physical Quantities and Measurement

A block of me...

Question

A block of metal of mass $0.20 k g$ is heated to $150^{\circ} C$ and dropped in copper calorimeter of water equivalent $0.025 k g$ , containing $150 g$ of water at $27^{\circ} C$ . The final temperature is $40^{\circ} C$ . Find the specific heat of the metal.
Given, specific heat of water $= 4.2 \times 10^{3} J - k g^{- 1} K^{- 1}$ .

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Solution

Mass of water $(m_{1}) = 150 g = 0.15 g$
Mass of metal $(m_{2}) = 0.20 k g$
Water equivalent of calorimeter $W = 0.025 k g$
$t_{1} = 27^{\circ} C, t_{2} = 150^{\circ} C, t = 40^{\circ} C, s_{1} = 4.2 \times 10^{3} J k g^{- 1} K^{- 1}$
Heat gained by water and calorimeter $= (m_{1} + w) \times s_{1} \times (t - t_{1})$
$= (0.15 + 0.025) \times 4.2 \times 10^{3} \times (40 - 27) = 9.555 \times 10^{3} J$
Heat lost by metal $= m_{2} \times s \times (t_{2} - t) = 0.20 \times s \times (150 - 40) = 22 s J$ , where $s$ is the specific heat of metal,
Equating $(1)$ and $(2)$ , we have
$s = 0.43 \times 10^{3} J k g^{- 1} K^{- 1}$ .

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