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(B) The mass of $1\, L$ of water is $m = 1\, kg = 1000\, g$.The specific heat capacity of water is $s = 4200\, J/(kg\cdot ^oC)$.The change in temperat

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$2\, min\, 48\, sec$

Vedclass · Answer

(B) The mass of $1\, L$ of water is $m = 1\, kg = 1000\, g$.The specific heat capacity of water is $s = 4200\, J/(kg\cdot ^oC)$.The change in temperature is $\Delta 	heta = 60\, ^oC - 20\, ^oC = 40\, ^oC$.The heat energy required is $Q = m s \Delta 	heta$.$Q = 1\, kg 	imes 4200\, J/(kg\cdot ^oC) 	imes 40\, ^oC = 168000\, J$.The power of the heater is $P = 1000\, W$.The time required is $t = Q / P$.$t = 168000\, J / 1000\, W = 168\, s$.Converting $168\, s$ into minutes and seconds: $168\, s = 2\, min\, 48\, s$.

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

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