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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: i. the heat capacity of 4.0 kg of liquid, ii. the specific heat capacity of the

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प्रश्न

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:

  1. the heat capacity of 4.0 kg of liquid,
  2. the specific heat capacity of the liquid.
संख्यात्मक
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उत्तर

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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अध्याय 11: Calorimetry - EXERCISE-11(A) [पृष्ठ २७१]

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सेलिना Concise Physics [English] Class 10 ICSE
अध्याय 11 Calorimetry
EXERCISE-11(A) | Q 7. | पृष्ठ २७१

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