Question

How many grams of a liquid of specific heat 0.2 at a temperature 40°C must be mixed with 100 gm of a liquid of specific heat of 0.5 at a temperature 20°C, so that the final temperature of the mixture becomes 32°C

• A. 175 gm
• B. 300 g
• C. 295 gm
• D. 375 g

Answer 1

Answer: (D)

375 g

Answer 2

Temperature of mixture

$\theta = \frac{m_{1}c_{1}\theta_{1} + m_{2}c_{2}\theta_{2}}{m_{1}c_{1} + m_{2}\theta_{2}}$

⇒ $32 = \frac{m_{1} \times 0.2 \times 40 + 100 \times 0.5 \times 20}{m_{1} \times 0.2 + 100 \times 0.5}$⇒ $m_{1} = 375gm$