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Standard IX

Chemistry

The Ideal Gas Equation

1.0 kg of wat...

Question

1.0 kg of water is contained in a 1.25 kW kettle. Calculate the time taken for the temperature of water to rise from $25^{\circ} C$ to its boiling point $100^{\circ} C$ . Specific heat capacity of water = $4.2 J g^{- 1} K^{- 1}$ .

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Solution

Let the time taken is t
Heat loss by kettle=heat gain water
$1.25 \times 10^{3} \times t = Q = m C Δ T$
C is the specific heat capacity of water
eg: C= $42$ J/kg/k
$Δ T = (100 - 25) ℃$
$1250 \times t = 1 \times 4200 \times (100 - 25) 1250 \times t = 4200 \times 75 50 t = 12600 t = \frac{1260}{5} = 252$ sec

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1.0 kg of water is contained in a 1.25 kW kettle. Calculate the time taken for the temperature of water to rise from 25∘C to its boiling point 100∘C. Specific heat capacity of water =4.2 J g−1K−1 .

Let the time taken is tHeat loss by kettle=heat gain water1.25×103×t=Q=mCΔTC is the specific heat capacity of watereg: C=42J/kg/kΔT=(100−25)℃1250×t=1×4200×(100−25)1250×t=4200×7550t=12600t=12605=252sec

1.0 kg of water is contained in a 1.25 kW kettle. Calculate the time taken for the temperature of water to rise from $25^{\circ} C$ to its boiling point $100^{\circ} C$ . Specific heat capacity of water = $4.2 J g^{- 1} K^{- 1}$ .