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.

Other
ANSWERED
asked 2021-03-15
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.

Answers (1)

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
0
 
Best answer

expert advice

Have a similar question?
We can deal with it in 3 hours
...