How many grams of ice at -16°C must be added to 168 grams of water that is initi
ID: 1478959 • Letter: H
Question
How many grams of ice at -16°C must be added to 168 grams of water that is initially at a temperature of 85°C to produce water at a final temperature of 13°C? Assume that no heat is lost to the surroundings and that the container has negligible mass. The specific heat of liquid water is 4190 J/kg °C and of ice is 2000 J/kg °C. For water the normal melting point is 0°C and the heat of fusion is 334 it × 103 J/kg. The normal boiling point is 100° C and the heat of vaporization is 2.256 × 106 J/kg.
Explanation / Answer
let x gram of ICE
ENERGY REQUIRED BY ice = mL + mS del T
= x * 334 + x * 2 * 16
energy lost by water = mS delT
= 168* 4.190 (85-13)
both must be equal
x * 334 + x * 2 * 16 = 168* 4.190 (85-13)
x= 138.476065574 gm
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