A solid of mass $50$ g at $150^{o}$ C is placed in $100$ g of water at $11^{o}$ C, when the final temperature recorded is $20^{o}$ C. Find the specific heat capacity of the solid. (Specific heat capacity of water $4.2$ J $/ g^{o}$ C

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Solution

Mass of solid $m_{s} = 50 g$
Temperature of solid $T_{s} = 150^{o} C$
Let the Specific heat capacity of solid be $S_{s}$ .
Mass of water $m_{w} = 100 g$
Temperature of water $T_{w} = 11^{o} C$
Specific heat capacity of water $S_{w} = 4.2 J / g^{o} C$
Final temperature $T_{f} = 20^{o} C$ .
Heat released by the solid is equal to the heat absorbed by water in order to achieve equilibrium.
$∴$ $m_{s} S_{s} (T_{s} - T_{f}) = m_{w} S_{w} (T_{f} - T_{w})$
Or $50 \times S_{s} (150 - 20) = 150 \times 4.2 \times (20 - 11)$
$⟹ S_{s} = 0.87 J / g^{o} C$

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A solid of mass 50g at 150oC is placed in 100g of water at 11oC, when the final temperature recorded is 20oC. Find the specific heat capacity of the solid. (Specific heat capacity of water 4.2J/goC

A solid of mass $50$ g at $150^{o}$ C is placed in $100$ g of water at $11^{o}$ C, when the final temperature recorded is $20^{o}$ C. Find the specific heat capacity of the solid. (Specific heat capacity of water $4.2$ J $/ g^{o}$ C