A piece of metal weighing 100 ,g is heated to | vedclass

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(C) The principle of calorimetry states that heat lost by the hot body equals heat gained by the cold body.Heat lost by metal = Heat gained by water.$

Vedclass · Accepted Answer

$0.45$

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(C) The principle of calorimetry states that heat lost by the hot body equals heat gained by the cold body.Heat lost by metal = Heat gained by water.$m_{	ext{metal}} 	imes s_{	ext{metal}} 	imes \Delta T_{	ext{metal}} = m_{	ext{water}} 	imes s_{	ext{water}} 	imes \Delta T_{	ext{water}}$Given: $m_{	ext{metal}} = 100 \,g = 0.1 \,kg$,$T_{	ext{initial, metal}} = 80^{\circ} C$,$T_{	ext{final}} = 15.69^{\circ} C$,$m_{	ext{water}} = 1 \,kg = 1000 \,g$,$T_{	ext{initial, water}} = 15^{\circ} C$,$s_{	ext{water}} = 4.18 \,J/g^{\circ} C$.Substituting the values:$100 	imes s_{	ext{metal}} 	imes (80 - 15.69) = 1000 	imes 4.18 	imes (15.69 - 15)$$100 	imes s_{	ext{metal}} 	imes 64.31 = 1000 	imes 4.18 	imes 0.69$$6431 	imes s_{	ext{metal}} = 2884.2$$s_{	ext{metal}} = \frac{2884.2}{6431} \approx 0.4485 \,J/g^{\circ} C \approx 0.45 \,J/g^{\circ} C$.

$A$ piece of metal weighing $100 \,g$ is heated to $80^{\circ} C$ and dropped into $1 \,kg$ of cold water in an insulated container at $15^{\circ} C$. If the final temperature of the water in the container is $15.69^{\circ} C$,the specific heat of the metal in $J / g^{\circ} C$ is:

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