JEE PYQ: Thermodynamics Question 44
Question 44 - 2019 (11 Jan Shift 2)
A metal ball of mass 0.1 kg is heated upto 500$°$C 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 30$°$C. What is the approximate percentage increment in the temperature of the water? [Specific Heat Capacities of water and metal are, respectively, 4200 Jkg$^{-1}$K$^{-1}$ and 400 Jkg$^{-1}$K$^{-1}$]
(1) 15%
(2) 30%
(3) 25%
(4) 20%
Show Answer
Answer: (4)
Solution
Assume final temperature $= T°$C
Heat loss = Heat gain
$m_{\text{metal}} s_{\text{metal}} \Delta T_M = m_w s_w \Delta T_w$
$0.1 \times 400 \times (500 - T) = 0.5 \times 4200 \times (T - 30) + 800(T - 30)$
$40(500 - T) = (T - 30)(2100 + 800)$
$20000 - 40T = 2900T - 30 \times 2900$
$20000 + 30 \times 2900 = T(2940)$
$T = 30.4°$C
$\frac{\Delta T}{T} \times 100 = \frac{6.4}{30} \times 100 = 21%$
So the closest answer is 20%.
BETA
