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10-1.Thermometry, Thermal Expansion and Calorimetry
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If an electric heater is rated at $1000\,W$, then the time required to heat one litre of water from $20\,^oC$ to $60\,^oC$ is
A
$1\,min\,\, 24\,sec$
B
$2\,min\,\, 48\,sec$
C
$4\,min\,\, 17\,sec$
D
$5\,min\,\, 36\,sec$
Solution
Heat required to heat $1 \mathrm{Lt}$
water $=m s \Delta \theta$
$=1 \times 4200 \times(60-20)$
$=4200 \times 40$
requred time $=\frac{4200 \times 40}{1000}$
$=42 \times 4=168$ second
i.e. $2$ min $48$ second
Standard 11
Physics
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