Question

An electric immersion heater is switched on for 8 minutes. The heat supplied by it raises the temperature of 500 g of water from $10{}^{o}Cto60{}^{o}C$. Calculate the power of the heater in watts.

Answer 1

$Time=8min.=480s$
$Letpowerofheater=P$
$Energysuppliedbyheater=P\times t$
$=P\times 480s$
Also, rise in. temp . of water ${\theta }_R$
$=(60-10)$
$=50{}^{o}C$
$\therefore$ Heat energy absorbed by water
$=mc{\theta }_R$
$=500g\times 4.2J{g}^{-1}{}^o{C}^{-1}\times 50{}^{o}C$
$=105000J$
$\therefore$ By the law of conservation of energy,
$P\times 480s=105000J$
$\therefore P=\frac{105000}{480}=219J{s}^{-1}orwatt$