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The specific heat of ice is 2.108 J/(g \'C), the specific heat of water is 4.184

ID: 933250 • Letter: T

Question

The specific heat of ice is 2.108 J/(g 'C), the specific heat of water is 4.184 J/(g ' C).
How much heat is absorbed by 100.0 g of ice as its temperature increases from -35.6 'C to -15.4 'C?
A.42.6 J B. 4.260 J C. 8.450 J D. 10.800 J
* Please explain how you got your answer. Much appreciated . The specific heat of ice is 2.108 J/(g 'C), the specific heat of water is 4.184 J/(g ' C).
How much heat is absorbed by 100.0 g of ice as its temperature increases from -35.6 'C to -15.4 'C?
A.42.6 J B. 4.260 J C. 8.450 J D. 10.800 J
* Please explain how you got your answer. Much appreciated .
How much heat is absorbed by 100.0 g of ice as its temperature increases from -35.6 'C to -15.4 'C?
A.42.6 J B. 4.260 J C. 8.450 J D. 10.800 J
* Please explain how you got your answer. Much appreciated .

Explanation / Answer

answer : B )  4260 J

mass (m) = 100 g

Cp = 2.108 J / g oC

T1 = -35.6 'C

T2 = -15.4 'C

dT = T2 -T1

= (-15.4 + 35.6)

= 20.2 oC

Q =m Cp dT

Q = 100 x 2.108 x 20.2

Q = 4258.16 J

Q = 4260 J

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