Three 45g ice cubes at 0 degrees Celsius are dropped into 5.00x10^2 mL of tea to make ice tea. The tea was initially at 20.0 degrees Celsius; when thermal equilibrium was reached, the final temperature was 0 degrees Celsius. How much of the ice melted and how much remained floating in the beverage? Assume the specific heat capacity of tea is the same as that of pure water.
Please show work. Thanks a lot!
2007-10-11 15:04:09 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
We need some more data to solve this problem.
Enthalpy of fusion (symbol: ΔHfus), also known as the heat of fusion or specific melting heat, of water, is 79.72cal/g.
Specific heat capacity of pure water: 1cal/g·C.
Density of water: 1g/mL
OK. That is enough. The rest is algebra.
Let X grams of ice to melt to cool the tea to oC.
(X g)*79.72cal/g = (5.00x10^2 mL)*(1g/mL)*(1cal/g·C)*(20C)
X = 125(g)
So ice melted: 125g
ice left: 3x45 - 125 = 10g
2007-10-13 07:34:34 · answer #1 · answered by Hahaha 7 · 1⤊ 0⤋