18.46 74. Calculate the amount of heat that must be absorbed by 10.0 g of ice at -20°C to convert it to liquid water at 60.0°C, Given: specific heat (ice) 2.1 J/g.°C; specific heat (water) = 4.18 J/g.°C; AHfus = 6.0 kJ/mol. A. 63 kJ (m. c.DT) + (n DH)+ (m.e+AT) ka fce tohan B. 7.5 J Cac.07ee= (log(21g:) 420 53り %3D C. 420 J D. 2.900 ( DA)- (6.65 S(6.0 kJ) =3:33 E. 6,300 J 14

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18.06
74.
Calculate the amount of heat that must be absorbed by 10.0 g of ice at -20°C to convert it to liquid water at 60.0°C, Given:
specific heat (ice) 2.1 J/g.°C; specific heat (water) = 4.18 J/g.°C; AHfus = 6.0 kJ/mol.
A. 63 kJ (m. c.DT) + (n DH)+ (m.e+AT)ka
fce
tohan
tus
B. 7.5 J
5313
%3D
C. 420 J
D. 2.900 ( DA)- (6. 65 S (6.0 kJ) =3-33
6,300 J
14
Transcribed Image Text:18.06 74. Calculate the amount of heat that must be absorbed by 10.0 g of ice at -20°C to convert it to liquid water at 60.0°C, Given: specific heat (ice) 2.1 J/g.°C; specific heat (water) = 4.18 J/g.°C; AHfus = 6.0 kJ/mol. A. 63 kJ (m. c.DT) + (n DH)+ (m.e+AT)ka fce tohan tus B. 7.5 J 5313 %3D C. 420 J D. 2.900 ( DA)- (6. 65 S (6.0 kJ) =3-33 6,300 J 14
Expert Solution
Step 1

The amount of energy that is transmitted as a result of the change in temperature from one system to its surroundings is called heat.

When heat is absorbed by the system, heat is positive (+)ve.

When heat is removed from the system, heat is negative (-)ve.

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