JEE MAIN - Physics (2004 - No. 19)
Time taken by a $$836$$ $$W$$ heater to heat one litre of water from $$10{}^ \circ C$$ to $$40{}^ \circ C$$ is
$$150$$ $$s$$
$$100$$ $$s$$
$$50$$ $$s$$
$$200$$ $$s$$
Explanation
$$\Delta Q = mC \times \Delta T$$
$$ = 1 \times 4180 \times \left( {40 - 10} \right) = 80 \times 30$$
( $$\therefore$$ $$\Delta Q = $$ heat supplied in time $$t$$ for heating $$1L$$ water from $${10^ \circ }C$$ to $${40^ \circ }C$$ )
also $$\Delta Q = 836 \times t \Rightarrow t = {{4180 \times 30} \over {836}} = 150\,s$$
$$ = 1 \times 4180 \times \left( {40 - 10} \right) = 80 \times 30$$
( $$\therefore$$ $$\Delta Q = $$ heat supplied in time $$t$$ for heating $$1L$$ water from $${10^ \circ }C$$ to $${40^ \circ }C$$ )
also $$\Delta Q = 836 \times t \Rightarrow t = {{4180 \times 30} \over {836}} = 150\,s$$
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