A 4.1-kg block of ice originally at 263 K is placed in thermal contact with a...
A 4.1-kg block of ice originally at 263 K is placed in thermal contact with a 13.1-kg block of silver (cAg = 233 J/kg-K) which is initially at 1064 K. The H2O - silver system is insulated so no heat flows into or out of it.
1) At what temperature will the system achieve equilibrium?
________K
Homework Answers
heat gained by ice = Mice*Cice*dT1 + mice*Lf +
mice*cw*dT2
Qgain = (4.1*2100*(273-263)) +
(4.1*334000)+(4.1*4190*(T-273))
Qgain = 1455500 + 17179T - 4689867
Qgain = 17179T - 3.234367e+6
heat lost by silver = mAg*cAg*dT3
Qloss = 13.2*233*(1064-T) = 3272438.4 -
3075.6T
from pricilple of calorimetry
Qlosss = Q gain
3272438.4 - 3075.6T = 17179T - 3.234367e+6
T = 321 K
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