JEE PYQ: Thermodynamics Question 18
Question 18 - 2020 (03 Sep Shift 2)
A calorimeter of water equivalent 20 g contains 180 g of water at 25$°$C. ‘$m$’ grams of steam at 100$°$C is mixed in it till the temperature of the mixture is 31$°$C. The value of ‘$m$’ is close to (Latent heat of water = 540 cal g$^{-1}$, specific heat of water = 1 cal g$^{-1}$ $°$C$^{-1}$)
(1) 2
(2) 4
(3) 3.2
(4) 2.6
Show Answer
Answer: (1)
Solution
Heat given by water $= m_w C_w (T_{\text{mix}} - T_w) = 200 \times 1 \times (31 - 25)$
Heat taken by steam $= m L_{\text{steam}} + m C_w (T_s - T_{\text{mix}})$
$= m \times 540 + m(1)(100 - 31)$
$= m \times 540 + m(1)(69)$
From the principle of calorimeter, Heat lost = Heat gained
$(200)(31 - 25) = m \times 540 + m(1)(69)$
$1200 = m(609) \Rightarrow m \approx 2$
BETA
