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

333
750
375
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Solution

The correct option is B 750
Given:
Specific heat capacity of Liquid A : $0.84 J g^{- 1} k^{- 1}$
Temperature of Liquid A : $50^{\circ} C$
Specific heat capacity of Liquid B : $2.1 J g^{- 1} k^{- 1}$
Temperature of Liquid B : $30^{\circ} C$
Mass of Liquid B : $200 g$

Change in temperature of liquid A:
$Δ T_{A} = 42 - 50 = - 8^{\circ} C$
Change in temperature of liquid B:
$Δ T_{B} = 42 - 30 = 12^{\circ} C$

Let $m$ be the mass of liquid A.

Heat energy lost by liquid A:
$H_{1} = m \times 0.84 \times - 8$

Heat energy gained by $100 g$ of liquid B:
$H_{2} = 200 \times 2.1 \times (12)$
NOTE: negative and positive sign only represent heat given and taken,
Assuming there is no heat loss,
Heat energy lost by A = Heat energy gained by B
$H_{1} = H_{2}$
$m \times 0.84 \times 8 = 200 \times 2.1 \times 12$
$m = 750 g$

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