The temperature of a piece of aluminum of 20gm mass is 90°C. It is placed on ice. If ice melted is 50gm, the specific heat of aluminum in cal/gmºC is
a.1 b. 3.22 C. 2.22 d. 5.32 •
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Answer:
Container contains m
1
=100g of Al and m
2
=200g of ice
In 4 minutes heat gives to system, Q=rate×time
Q=100×4×60=24000cal
Assume the whole ice melts to water (thereby all mass of ice m
2
gains
( latent heat ) and change in temperature be ΔT, T→ final temp.
Thus, from heat-transfer law.
Q=m
1
S
1
ΔT+m
2
S
2
(20)+m
2
L+m
2
(1)T (water phase) [ice consume heat in −20
o
C to 0
o
C only]
=24000=100×0.2(T+20)+200×0.5×20+200×80+200T
=24000=20(T+20)+2000+16000+200T
=220T=5600
T=
22
560
=25.45
o
C
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