JEE PYQ: Thermodynamics Question 29
Question 29 - 2020 (07 Jan Shift 2)
M grams of steam at 100$°$C is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40$°$C [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g], the value of M is __________.
Show Answer
Answer: 40
Solution
Using the principal of calorimetry:
$M_{\text{ice}} L_f + m_{\text{ice}}(40 - 0)C_w = m_{\text{steam}} L_v + m_{\text{steam}}(100 - 40)C_w$
$M(540) + M \times 1 \times (100 - 40) = 200 \times 80 + 200 \times 1 \times 40$
$600M = 24000$
$M = 40$ g
BETA
