Physics, asked by dipshikabhattarai671, 5 hours ago

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 • ​

Answers

Answered by Kiranpagare
0

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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