Ncert Exercises & Solutions

11.14 In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 ${cm}^{3}$ of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for the specific heat of the metal?

Water equivalent of calorimeter, m’ = 0.025 kg = 25 g

Volume of water, V = 150 cm $^{3}$

Mass (M) of water at temperature 27°C = 150 × 1 = 150 g

Fall in the temperature of the metal:

ΔT = T $_{1}$ – $T_{2}$ = 150 – 40 = 110°C

Specific heat of water, $c_{w}$ = 4.186 J/g-°K

If the specific heat of the metal = c,

Heat lost by the metal, θ = mcΔT

Heat gained by the water and calorimeter system:

$Δθ' = m_{1} c_{w} ΔT ’ = (M + m') c_{w} ΔT ’$

Heat lost by the metal = Heat gained by the water and calorimeter system

$mcΔT = (M + m ’) c_{w} ΔT ’$

200 × c × 110 = (150 + 25) × 4.186 × 13

$∴ c = \frac{175 \times 4.186 \times 13}{110 \times 20} = 0.43 J g^{- 1} K^{- 1}$

If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.

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