Thermal Properties of Matter
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Question : 14
Total: 22
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 cm3 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 specific heat of the metal?
Solution:
Here, mass of metal, m = 0.20 k g = 200 g
Fall in temperature of metal
Δ T = 150 – 40 = 110 ° C
If c is specific heat of the metal, then heat lost by the metal
,Δ Q = m c D T = 200 × c × 110
Volume of water= 150 c . c .
∴ Mass of water , m ′ = 150 g
Water equivalent of calorimeter,w = 0.025 k g = 25 g
Rise in temperature of water and calorimeter,Δ T ′ = 40 – 27 = 13 ° C
Heat gained by water and calorimeter,
Δ Q ′ = ( m ′ + w ) Δ T ′
= ( 150 + 25 ) × 13 = 175 × 13
A s , Δ Q = Δ Q ′
∴ From (i) and (ii),200 × c × 110 = 175 × 13
or c =
≈ 0.1
If some heat is lost to the surroundings, value of c so obtained will be less than the actual value ofc .
Fall in temperature of metal
If c is specific heat of the metal, then heat lost by the metal
,
Volume of water
Water equivalent of calorimeter,
Rise in temperature of water and calorimeter,
Heat gained by water and calorimeter,
∴ From (i) and (ii),
or
If some heat is lost to the surroundings, value of c so obtained will be less than the actual value of
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