Problem 17.92 You cool a 120.0 g slug of red-hot iron (temperature 745 C ) by dr
ID: 1568731 • Letter: P
Question
Problem 17.92
You cool a 120.0 g slug of red-hot iron (temperature 745 C ) by dropping it into an insulated cup of negligible mass containing 65.0 g of water at 20.0 C . Assume no heat exchange with the surroundings.
Part A
What is the final temperature of the water?
Express your answer using three significant figures.
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Part B
What is the final mass of the iron and the remaining water?
Express your answer using three significant figures.
145
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Correct
ANSWER 1: Deduction: -3%
ANSWER 2: Deduction: -3%
145.48
ANSWER 3: Deduction: -3%
144.6
145
g
These are wrong for part A
Problem 17.92
You cool a 120.0 g slug of red-hot iron (temperature 745 C ) by dropping it into an insulated cup of negligible mass containing 65.0 g of water at 20.0 C . Assume no heat exchange with the surroundings.
Part A
What is the final temperature of the water?
Express your answer using three significant figures.
SubmitHintsMy AnswersGive UpReview Part
Incorrect; Try Again; 3 attempts remaining
Part B
What is the final mass of the iron and the remaining water?
Express your answer using three significant figures.
Tfinal =145
CSubmitHintsMy AnswersGive UpReview Part
Correct
ANSWER 1: Deduction: -3%
ANSWER 2: Deduction: -3%
Tfinal =145.48
CANSWER 3: Deduction: -3%
Tfinal =144.6
C Tfinal =145
Cg
These are wrong for part A
Explanation / Answer
C(Iron) = 0.460 j/gc
C(water) = 4.19 j/gc
L(water) = 2264.76
m(iron) = 120g ; m(water) = 65g ; Tw = 20 c; ti = 745c
Q(lost by slug) = Q(gained by water)
120x0.460x(745-100) = 65x4.19x(100-20)+m'x2264.76
35604 = 21788+2264.76m'
13816/2264.76 = m'
m' = 6.100 grams [ mass of water in vapour form]
PartA)
Final temp of water = 100 °C
Part B)
Mass of Iron remains same
Mass of remaing water = 65 -6.1 = 58.9 grams