A liquid of mass $100 g$ at $120 ° C$ is poured in water at $20 ° C$ , when the final temperature recorded is $40 ° C$ . If the specific heat capacity of the liquid is $0.8 J g^{- 1} ° C^{- 1}$ , calculate the initial mass of water.

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Solution

Step 1: Organised table of the given data
Substance Mass S.H.C Initial Temperature Final Temperature=x
Liquid $100 g$ $0.8 J g^{- 1} ° C^{- 1}$ $120 ° C$ $θ_{F} = 120 - 40 = 80 ° C$
Water $x$ $4.2 J g^{- 1} K^{- 1}$ $20 ° C$ $θ_{R} = 40 - 20 = 20 ° C$
Step 2: Calculating the heat energy
The heat released by the liquid is given as
$m c θ_{F} = 100 \times 0.8 \times 80 = 6400$
The heat absorbed by the water is given as
$m c θ_{R} = x \times 4.2 \times 20 = 84 x$
Step3: Calculating the initial mass
The heat released by liquid= Heat absorbed by water
$84 x = 6400 x = 76.19 g$
Hence the initial mass of the water is $76.19 g$

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Substance	Mass	S.H.C	Initial Temperature	Final Temperature=x
Liquid	$100 g$	$0.8 J g^{- 1} ° C^{- 1}$	$120 ° C$	$θ_{F} = 120 - 40 = 80 ° C$
Water	$x$	$4.2 J g^{- 1} K^{- 1}$	$20 ° C$	$θ_{R} = 40 - 20 = 20 ° C$

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