A small electric heater is used to heat 200 , | vedclass

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(D) Given: Mass of water $m = 200 \,g = 0.2 \,kg$, time $t = 200 \,s$, specific heat $s = 4200 \,J \,kg^{-1} \,K^{-1}$, and temperature change $\Delta

Vedclass · Accepted Answer

$252$

Vedclass · Answer

(D) Given: Mass of water $m = 200 \,g = 0.2 \,kg$, time $t = 200 \,s$, specific heat $s = 4200 \,J \,kg^{-1} \,K^{-1}$, and temperature change $\Delta T = 100^{\circ} C - 40^{\circ} C = 60 \,K$.
The heat energy $Q$ required to raise the temperature of the water is given by $Q = m \cdot s \cdot \Delta T$.
The power $P$ supplied by the heater is $P = \frac{Q}{t} = \frac{m \cdot s \cdot \Delta T}{t}$.
Substituting the values: $P = \frac{0.2 \,kg 	imes 4200 \,J \,kg^{-1} \,K^{-1} 	imes 60 \,K}{200 \,s}$.
$P = \frac{50400}{200} \,W = 252 \,W$.

$A$ small electric heater is used to heat $200 \,g$ of water. The time required to bring all this water from $40^{\circ} C$ to $100^{\circ} C$ is $200 \,s$. If the specific heat of the water is $4200 \,J \,kg^{-1} \,K^{-1}$, then the power supplied by the heater is: (in $\,W$)

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