A solid of mass $25 g$ (sp. heat capacity $0.8 J / g ° C$ ) and at $120 ° C$ is placed in $100 g$ of water at $20 ° C$ . Calculate the final temperature of the mixture. Specific heat capacity of water is $4.2 J / g^{- 1} ° C$ .

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Solution

Step 1: Given data
Solid: Mass $(m) = 25 g$
specific heat capacity $(c) = 0.8 J / g^{- 1} ° C$
Initial temperature $= 120 ° C$
Water: mass $(m) = 100 g$
specific heat capacity $(c) = 4.2 J / g^{- 1} ° C$
Initial temperature $= 20 ° C$
Let the final temperature be $x$ .
Step 2: calculating the heat absorbed and given out
The change in temperature of solid $(θ_{F}) = 120 ° C - x$
The change in temperature of water $(θ_{R}) = x - 20 ° C$
Now, heat is given out by the solid=heat absorbed by the water
$\Rightarrow m c θ_{F} = m c θ_{R} \Rightarrow 25 \times 0.8 \times (120 - x) = 100 \times 4.2 \times (x - 20) \Rightarrow 2400 - 20 x = 420 x - 8400 \Rightarrow 440 x = 10800 \Rightarrow x = \frac{10800}{440} \Rightarrow x = 24.54 ° C$
The final temperature of the mixture is $24.54 ° C$ .

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A solid of mass 25g(sp. heat capacity 0.8J/g°C) and at 120°C is placed in 100gof water at 20°C. Calculate the final temperature of the mixture. Specific heat capacity of water is 4.2J/g-1°C.

A solid of mass $25 g$ (sp. heat capacity $0.8 J / g ° C$ ) and at $120 ° C$ is placed in $100 g$ of water at $20 ° C$ . Calculate the final temperature of the mixture. Specific heat capacity of water is $4.2 J / g^{- 1} ° C$ .