1. The following scenarios apply to the process of calorimetry as a means to qua
ID: 1838713 • Letter: 1
Question
1. The following scenarios apply to the process of calorimetry as a means to quantitatively measure heat transfer.
a. How many grams of ice at -17 °C must be added to 741 g of water that is initially at a temperature of 70 °C to produce water at a final temperature of 12 °C? Assume that no heat is lost to the surroundings and that the container has negligible mass. The specific heat of liquid water is 4,190 J/kg °C and that of ice is 2,000 J/kg °C. For water, the normal melting point is 0 °C, and the heat of fusion is 334 × 103 J/kg. The normal boiling point is 100 °C, and the heat of vaporization is 2.256 × 106 J/kg.
b. An unknown solid sample with a mass of 85.0 g at a temperature of 100.0 oC is placed in a calorimeter. The calorimeter vessel is made of 0.1500 kg of copper and contains 0.200 kg of water. Both the calorimeter vessel and the water are initially at 19.0 oC.
The final temperature of the system is measured to be 26.1 oC
What is the specific heat of the sample? (Assume no heat loss to the surroundings.)
The specific heat of copper is 390 J/kg-K.
The specific heat of water is 4.19E3 J/kg-K.
Explanation / Answer
Heat absorbed by ice = m Cice deltaT + m Lf + m Cwater deltaT
= m x 2000 x 17 + m x 334 x 10^3 + m x 4190 x (12 - 0)
= 418280m
Heat released by water = 0.741 kg x 4190 J/kg C x (70 - 12)
= 180077.82 J
As there is no heat loss then
418280m = 180077.82
m = 0.430 kg Or 430 grams
(B) heat loss by sample = m C delta
= 0.085 x C x (100 - 26.1)
= 6.28C
Heat gain = 0.1500 x 390 x (26.1 - 19) + 0.200 x 4190 x (26.1 - 19)
= 6365.15 J
as there is no loss to surrouding.
6.28 C = 6365.15
C= 1013.56 J / kg deg C
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