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

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

Tardigrade Admin · Accepted Answer

Temperature of mixture θ=(m1 c1 θ1+m2 c2 θ2/m1 c1+m2 θ2) ⇒ 32=(m1 × 0.2 × 40+100 × 0.5 × 20/m1 × 0.2+100 × 0.5) ⇒ m1=375 gm

Q. How many grams of a liquid of specific heat $0.2$ at a temperature $4 0^{\circ} C$ must be mixed with $100 g m$ of a liquid of specific heat of $0.5$ at a temperature $2 0^{\circ} C$ , so that the final temperature of the mixture becomes $3 2^{\circ} C$

175 gm

300 g

295 gm

375 g

Temperature of mixture
$θ = \frac{m _{1} c _{1} θ _{1} + m _{2} c _{2} θ _{2}}{m _{1} c _{1} + m _{2} θ _{2}}$
$\Rightarrow 32 = \frac{m _{1} \times 0.2 \times 40 + 100 \times 0.5 \times 20}{m _{1} \times 0.2 + 100 \times 0.5}$
$\Rightarrow m_{1} = 375 g m$

Q. How many grams of a liquid of specific heat 0.2 at a temperature 40∘C must be mixed with 100gm 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