A 20g piece of iron (specific heat capacity =0.5 J/g∘C ) at 200∘C is immersed in 100g water of 27∘C . What will be the temperature ∘C) of this system of thermal equilibrium

At equilibrium Heat lost by iron = Heat gain by water20×0.5×(473-T) =100×4.2×(T-300) T =304K=31∘C

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A 20g piece of iron (specific heat capacity $= 0.5 J / g^{\circ} C$ ) at $200^{\circ} C$ is immersed in 100g water of $27^{\circ} C$ . What will be the temperature $(^{\circ} C)$ of this system of thermal equilibrium

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answer is 31.

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Detailed Solution

At equilibrium Heat lost by iron = Heat gain by water

$\begin{array}{rrr} 20 \times 0 .5 \times (473 - T) & = 100 \times 4 .2 \times (T - 300) \\ \begin{array}{rr} T & = 304 K = 31^{\circ} C \end{array} \end{array}$

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