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.

pancha3 2021-03-15 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.
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Expert Answer

Roosevelt Houghton
Answered 2021-03-17 Author has 106 answers
Mass of ice (mi)=0.018 kg
Latent Heat of ice ( Li ) = 333000 J / kg
Total heat for the melting of ice Qi=miLi
= 5994.0 J
Mass of Lead (mL) = 0.75 kg
Temp. of Lead (tL) = 255 C
Spec.heat capacity of Lead (cL)= 128 J / kg - K
Total heat comes from Lead QL=(mL)(cL)255
=24480 J
Since QL>Qi
So all ice will melt and the temperature will increase up to t C
Mass of water (mw) = 0.16 kg
Temp. difference of water(t)=t C
spec.heat capacity ofwater (Cw) = 4190 J / kg- K
Mass of copper (mcu)=0.100kg
Temp. difference of copper (t)=t C
spec.heat capacity of copper (Ccu) = 386 J / kg-K
Apply:
Principle of calometry
miLi+(micw+mwcw+mcuccu)(t)=mLcL(225t)
=5994.0J+(75.42+670.4+38.6)(t)=96(225t)
=5994.0J+784.42t=2160096t
216005994=(784.42+96)t
t=17.7 C
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