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

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(C) The heat lost by the metal is equal to the heat gained by the water in an insulated container.Let $c$ be the specific heat of the metal.Mass of me

Vedclass · Accepted Answer

$0.45$

Vedclass · Answer

(C) The heat lost by the metal is equal to the heat gained by the water in an insulated container.Let $c$ be the specific heat of the metal.Mass of metal $m_m = 100 \, g$,Initial temperature $T_{m,i} = 80^{\circ} C$.Mass of water $m_w = 1000 \, g$,Initial temperature $T_{w,i} = 15^{\circ} C$.Final equilibrium temperature $T_f = 15.69^{\circ} C$.Specific heat of water $c_w = 4.184 \, J / g \cdot ^{\circ} C$.Heat lost by metal $= m_m 	imes c 	imes (T_{m,i} - T_f) = 100 	imes c 	imes (80 - 15.69) = 6431c$.Heat gained by water $= m_w 	imes c_w 	imes (T_f - T_{w,i}) = 1000 	imes 4.184 	imes (15.69 - 15) = 4184 	imes 0.69 = 2886.96 \, J$.Equating the two: $6431c = 2886.96$.$c = \frac{2886.96}{6431} \approx 0.4489 \, J / g \cdot ^{\circ} C$.Rounding to two decimal places,we get $0.45 \, J / g \cdot ^{\circ} C$.

$A$ 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 $J / g \cdot ^{\circ} C$ is

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