Asked by JOHN
A 55.0 g ice cube, initially at 0°C, is dropped into a Styrofoam cup containing 318 g of water, initially at 18.2°C. What is the final temperature of the water, if no heat is transferred to the Styrofoam or the surroundings?
Answers
Answered by
Elena
Heat of fusion is the amount of heat energy required to change the state of a substance from solid to liquid.
λ = 335000 J/kg is heat of fusion of water-ice
c = 4185.5 J/(kg•K) is heat capacity of water
Q1 = λ•m1
Q2 =c•m1 •(t-t1) =c•m1•t
Q3 = c•m2 •(t2-t) .
Q1+Q2 = Q3
λ•m1 + c•m1•t = c•m2 •(t2-t),
λ•m1 + c•m1•t = c•m2 •t2 - c•m2 •t,
c•m1•t+ c•m2 •t = c•m2 •t2 - λ•m1
t = (c•m2•t2 - λ•m1)/[c•(m1+m2)] =( 4 180•0.318•18.2 -335000•0.055)/[4180•(0.335•0.055)] = 3.7 oC
λ = 335000 J/kg is heat of fusion of water-ice
c = 4185.5 J/(kg•K) is heat capacity of water
Q1 = λ•m1
Q2 =c•m1 •(t-t1) =c•m1•t
Q3 = c•m2 •(t2-t) .
Q1+Q2 = Q3
λ•m1 + c•m1•t = c•m2 •(t2-t),
λ•m1 + c•m1•t = c•m2 •t2 - c•m2 •t,
c•m1•t+ c•m2 •t = c•m2 •t2 - λ•m1
t = (c•m2•t2 - λ•m1)/[c•(m1+m2)] =( 4 180•0.318•18.2 -335000•0.055)/[4180•(0.335•0.055)] = 3.7 oC
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