JEE MAIN - Physics (2019 - 11th January Evening Slot - No. 22)
A metal ball of mass 0.1 kg is heated upto 500oC and dropped into a vessel of heat capacity 800 JK–1 and containing 0.5 kg water. The initial temperature of water and vessel is 30oC. What is the approximate percentage increment in the temperature of the water? [Specific Heat Capacities of water and metal are, respectively, 4200 Jkg–1 and 400 Jkg–1 K–1
20%
25%
15%
30%
Explanation
0.1 $$ \times $$ 400 $$ \times $$ (500 $$-$$ T) = 0.5 $$ \times $$ 4200 $$ \times $$ (T $$-$$ 30) + 800 (T $$-$$ 30)
$$ \Rightarrow $$ 40(500 $$-$$ T) = (T $$-$$ 30) (2100 + 800)
$$ \Rightarrow $$ 20000 $$-$$ 40T = 2900 T $$-$$ 30 $$ \times $$ 2900
$$ \Rightarrow $$ 20000 + 30 $$ \times $$ 2900 = T(2940)
T = 30.4oC
$${{\Delta T} \over T} \times 100$$ = $${{6.4} \over {30}} \times 100$$
$$ \simeq $$ 20%
$$ \Rightarrow $$ 40(500 $$-$$ T) = (T $$-$$ 30) (2100 + 800)
$$ \Rightarrow $$ 20000 $$-$$ 40T = 2900 T $$-$$ 30 $$ \times $$ 2900
$$ \Rightarrow $$ 20000 + 30 $$ \times $$ 2900 = T(2940)
T = 30.4oC
$${{\Delta T} \over T} \times 100$$ = $${{6.4} \over {30}} \times 100$$
$$ \simeq $$ 20%
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