Question

11. A peice of lead falls from a height of 8 metres. Calculate the rise in the temperature of the lead if 60% of heat generated is used up by the lead piece. (sp. heat of lead = 0.03 cal/gm°C]
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Answer 1

Answer:

Given,

h=100m

s=30.6cal/kg

0

C

Potential energy of lead= thermal energy of slab

mgh=msΔT

ΔT=

s

gh

ΔT=

30.6cal/kg

0

C

10×100

. . . . .(1)

we know that 1cal=4.2J

Equation (1) becomes,

ΔT=

30.6×4.2

1000

ΔT=7.62

0

C

Explanation:

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