JEE PYQ: Thermodynamics Question 40
Question 40 - 2019 (10 Jan Shift 2)
An unknown metal of mass 192 g heated to a temperature of 100$°$C was immersed into a brass calorimeter of mass 128 g containing 240 g of water at a temperature of 8.4$°$C. Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5$°$C. (Specific heat of brass is 394 J kg$^{-1}$ K$^{-1}$)
(1) 458 J kg$^{-1}$ K$^{-1}$
(2) 1232 J kg$^{-1}$ K$^{-1}$
(3) 916 J kg$^{-1}$ K$^{-1}$
(4) 654 J kg$^{-1}$ K$^{-1}$
Show Answer
Answer: (3)
Solution
Let specific heat of unknown metal be ‘$s$’. According to principle of calorimetry, Heat lost = Heat gain.
$m \times s \times \Delta\theta = m_{\text{brass}}(\Delta\theta_1 + m_2 s_{\text{water}} \times \Delta\theta_2)$
$192 \times S \times (100 - 21.5) = 128 \times 394 \times (21.5 - 8.4) + 240 \times 4200 \times (21.5 - 8.4)$
Solving we get, $S = 916$ J kg$^{-1}$ K$^{-1}$
BETA
