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The time required to raise the temperature of 3 litre of water from $0^{\circ} \mathrm{C}$ to $80^{\circ} \mathrm{C}$ by a heater operated under $200 \mathrm{~V}$ having resistance of $50 \Omega$ is
[specific heat capacity of water is $4200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ ] [density of water $=1000 \mathrm{~kg} / \mathrm{m}^3$ ]
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[specific heat capacity of water is $4200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ ] [density of water $=1000 \mathrm{~kg} / \mathrm{m}^3$ ]
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Verified Answer
The correct answer is:
$21 \mathrm{~min}$
Heat absorbed by water
$\Delta \mathrm{Q}_1=\mathrm{mS} \Delta \mathrm{T}$
$=(3 \times 4200 \times 80) \mathrm{J} \quad[\because \mathrm{m}=\mathrm{V} \rho]$
Also, heat generated by heater
$\Delta \mathrm{Q}_2=800 \times \mathrm{t} \quad\left[\because \mathrm{P}=\frac{\mathrm{V}^2}{\mathrm{R}}=\frac{200^2}{50}=800\right.$ watt $]$
So, $3 \times 4200 \times 80=800 \mathrm{t}$
$\mathrm{t}=1260 \mathrm{sec}=21 \mathrm{~min}$
$\Delta \mathrm{Q}_1=\mathrm{mS} \Delta \mathrm{T}$
$=(3 \times 4200 \times 80) \mathrm{J} \quad[\because \mathrm{m}=\mathrm{V} \rho]$
Also, heat generated by heater
$\Delta \mathrm{Q}_2=800 \times \mathrm{t} \quad\left[\because \mathrm{P}=\frac{\mathrm{V}^2}{\mathrm{R}}=\frac{200^2}{50}=800\right.$ watt $]$
So, $3 \times 4200 \times 80=800 \mathrm{t}$
$\mathrm{t}=1260 \mathrm{sec}=21 \mathrm{~min}$
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