# A copper calorimeter can with mass 0.100 kg contains 0.160 kgof water and 0.018 kg of ice in thermal equilibrium at atmospheric pressure. If 0.750 kg of lead at a temperature is dropped into the calorimeter can, what is thefinal temperature? Assume that no heat is lost to the surroundings.

A copper calorimeter can with mass 0.100 kg contains 0.160 kgof water and 0.018 kg of ice in thermal equilibrium at atmospheric pressure.
If 0.750 kg of lead at a temperature is dropped into the calorimeter can, what is thefinal temperature? Assume that no heat is lost to the surroundings.
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Roosevelt Houghton
Mass of ice (${m}_{i}$)=0.018 kg
Latent Heat of ice ( ${L}_{i}$ ) = 333000 J / kg
Total heat for the melting of ice ${Q}_{i}={m}_{i}\cdot {L}_{i}$
= 5994.0 J
Mass of Lead (${m}_{L}$) = 0.75 kg
Temp. of Lead (${t}_{L}$) = 255 ${C}^{\circ }$
Spec.heat capacity of Lead (${c}_{L}$)= 128 J / kg - K
Total heat comes from Lead ${Q}_{L}=\left({m}_{L}\right)\cdot \left({c}_{L}\right)\cdot 255$
=24480 J
Since ${Q}_{L}>{Q}_{i}$
So all ice will melt and the temperature will increase up to ${t}^{\circ }$ C
Mass of water (${m}_{w}$) = 0.16 kg
Temp. difference of water($\mathrm{△}t$)=t ${C}^{\circ }$
spec.heat capacity ofwater (${C}_{w}$) = 4190 J / kg- K
Mass of copper (${m}_{cu}$)=0.100kg
Temp. difference of copper ($\mathrm{△}t$)=t ${C}^{\circ }$
spec.heat capacity of copper (${C}_{cu}$) = 386 J / kg-K
Apply:
Principle of calometry
${m}_{i}{L}_{i}+\left({m}_{i}{c}_{w}+{m}_{w}{c}_{w}+{m}_{cu}{c}_{cu}\right)\left(t\right)={m}_{L}{c}_{L}\left(225-t\right)$
$=5994.0J+\left(75.42+670.4+38.6\right)\left(t\right)=96\left(225-t\right)$
$=5994.0J+784.42t=21600-96t$
$21600-5994=\left(784.42+96\right)t$
$t={17.7}^{\circ }$ C