Question #8832

Find the final temperature of a mixture of 200 grams of water at 85 degrees celsius placed inside an aluminum container of mass 20 grams into which is added 20 grams of ice at minus 10 degrees celsius. Please show formulas

Expert's answer

c(w)*(T(w)-T(f))*m(w)+c(ic)*(T(f)-T(ic))-l(ic)*m(ic)+c(al)*(T(al)-T(f))*m(al)=0,

(T(al)=T(w)), T(f)-final T =T

4200*0.2*(85-T)+2100*0.02*(T-10)

-335000*0.02+930*0.02*(85-Т)=0

(T(al)=T(w)), T(f)-final T =T

4200*0.2*(85-T)+2100*0.02*(T-10)

-335000*0.02+930*0.02*(85-Т)=0

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