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Question
An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10.0℃ to 15.0℃ in100 s. Calculate:
- the heat capacity of 4.0 kg of liquid,
- the specific heat capacity of the liquid.
Solution
Power of heater P = 600 W
t = 100 S
Mass of liquid m = 4.0 kg
Change in temperature of liquid = (15 − 10)°C = 5°C (or 5 K)
Time taken to raise its temperature = 100 s
(i) Heat capacity of liquid C' = `"Q"/(Δ"T")`
∴ C' = `60000/5`
= 12000 JK-1 or 1.2 × 104 JK-1
(ii) Specific heat capacity of liquid C = ?
C = `"Q"/("m"Δ"T")`
= `60000/(4 xx 5)`
= 3 × 103 J Kg-1°C-1
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