Question # 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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ANSWERED 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. 2021-03-17
Mass of ice ($$\displaystyle{m}_{{i}}$$)=0.018 kg
Latent Heat of ice ( $$\displaystyle{L}_{{i}}$$ ) = 333000 J / kg
Total heat for the melting of ice $$\displaystyle{Q}_{{i}}={m}_{{i}}\cdot{L}_{{i}}$$
= 5994.0 J
Mass of Lead ($$\displaystyle{m}_{{L}}$$) = 0.75 kg
Temp. of Lead ($$\displaystyle{t}_{{L}}$$) = 255 $$\displaystyle{C}^{\circ}$$
Spec.heat capacity of Lead ($$\displaystyle{c}_{{L}}$$)= 128 J / kg - K
Total heat comes from Lead $$\displaystyle{Q}_{{L}}={\left({m}_{{L}}\right)}\cdot{\left({c}_{{L}}\right)}\cdot{255}$$
=24480 J
Since $$\displaystyle{Q}_{{L}}{>}{Q}_{{i}}$$
So all ice will melt and the temperature will increase up to $$\displaystyle{t}^{\circ}$$ C
Mass of water ($$\displaystyle{m}_{{w}}$$) = 0.16 kg
Temp. difference of water($$\displaystyle\triangle{t}$$)=t $$\displaystyle{C}^{\circ}$$
spec.heat capacity ofwater ($$\displaystyle{C}_{{w}}$$) = 4190 J / kg- K
Mass of copper ($$\displaystyle{m}_{{{c}{u}}}$$)=0.100kg
Temp. difference of copper ($$\displaystyle\triangle{t}$$)=t $$\displaystyle{C}^{\circ}$$
spec.heat capacity of copper ($$\displaystyle{C}_{{{c}{u}}}$$) = 386 J / kg-K
Apply:
Principle of calometry
$$\displaystyle{m}_{{i}}{L}_{{i}}+{\left({m}_{{i}}{c}_{{w}}+{m}_{{w}}{c}_{{w}}+{m}_{{{c}{u}}}{c}_{{{c}{u}}}\right)}{\left({t}\right)}={m}_{{L}}{c}_{{L}}{\left({225}-{t}\right)}$$
$$\displaystyle={5994.0}{J}+{\left({75.42}+{670.4}+{38.6}\right)}{\left({t}\right)}={96}{\left({225}-{t}\right)}$$
$$\displaystyle={5994.0}{J}+{784.42}{t}={21600}-{96}{t}$$
$$\displaystyle{21600}-{5994}={\left({784.42}+{96}\right)}{t}$$
$$\displaystyle{t}={17.7}^{\circ}$$ C