Question 488: Thermodynamics & KTG

Question

Time taken by a $836$ $W$ heater to heat one litre of water from $10{}^ \circ C$ to $40{}^ \circ C$ is

Accepted Answer

$\Delta Q = mC 	imes \Delta T$
$ = 1 	imes 4180 	imes \left( {40 - 10} ight) = 80 	imes 30$
( $	herefore$ $\Delta Q = $ heat supplied in time $t$ for heating $1L$ water from ${10^ \circ }C$ to ${40^ \circ }C$ )
also $\Delta Q = 836 	imes t \Rightarrow t = {{4180 	imes 30} \over {836}} = 150\,s$

Answer

$150$ $s$

Answer

$100$ $s$

Answer

$50$ $s$

Answer

$200$ $s$

Thermodynamics & KTG

Time taken by a 836 W heater to heat one litre of water from 10{}^ \circ C to 40{}^ \circ C is

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