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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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