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

175 gm

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300 g

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295 gm

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375 g

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Solution

The correct option is B 375 g
Temperature of mixture $θ = \frac{m_{1} c_{1} θ_{1} + m_{2} c_{2} θ_{2}}{m_{1} c_{1} + m_{2} θ_{2}}$

$\Rightarrow 32 = \frac{m_{1} \times 0.2 \times 40 + 100 \times 0.5 \times 20}{m_{1} \times 0.2 + 100 \times 0.5} \Rightarrow m_{1} = 375 g m$

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Similar questions

Q. How many grams of a liquid of specific heat

0.2 {cal/g}^{\circ} C

at a temperature

40^{\circ} C

must be mixed with

100 gm

of a liquid of specific heat

0.5 {cal/g}^{\circ} C

at a temperature

20^{\circ} C

, so that the final temperature of the mixture becomes

32^{\circ} C

Liquid A of specific heat capacity 1 J g⁻¹ K⁻¹ at a temperature of $50^{\circ} C$ is mixed with 100g of a liquid B of specific heat capacity 2 J g⁻¹ K⁻¹ at $30^{\circ} C$ .The final temperature of mixture becomes $40^{\circ} C$ . Find the mass of liquid A.

Liquid A of specific heat capacity $0.84 J g^{- 1} k^{- 1}$ at a temperature of $50^{\circ} C$ is mixed with $200 g$ of a liquid B of specific heat capacity $2.1 J g^{- 1} k^{- 1}$ at $30^{\circ} C$ . The final temperature of mixture becomes $42^{\circ} C$ . The mass of liquid A (in $g$ ) will be