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Calculate the Time Required to Heat 20 Kg of Water from 10°C to 35°C Using an Immersion Heater Rated 1000 W.

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प्रश्न

Calculate the time required to heat 20 kg of water from 10°C to 35°C using an immersion heater rated 1000 W. Assume that 80% of the power input is used to heat the water. Specific heat capacity of water = 42000 J kg−1 K−1.

योग
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उत्तर

Given:-

Power rating of the immersion rod, P = 1000 W

Specific heat of water, S = 4200 J kg−1 K−1

Mass of water, M  = 20 kg

Change in temperature, ΔT = 25 °C

Total heat required to raise the temperature of 20 kg of water from 10°C to 35°C is given by

Q = M × S × ΔT

Q = 20 × 4200 × 25

Q = 20 × 4200 × 25 = 21 × 105 J

Let the time taken to heat 20 kg of water from 10°C to 35°C be t. Only 80% of the immersion rod's heat is useful for heating water. Thus,

Energy of the immersion rod utilised for heating the water = t × (0.80) × 1000 J

t × (0.80) × 1000 J  = 21 × 105 J

`t=(21xx10^5)/800=2625s`

`rArrt=2625/60=43.75minapprox44min`

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Anomalous Expansion of Water
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 25: Calorimetry - Exercises [पृष्ठ ४७]

APPEARS IN

एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 25 Calorimetry
Exercises | Q 8 | पृष्ठ ४७

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