# A piece of substance of specific heat capacity 450Jkg⁻¹k⁻¹ falls…

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A. 2/9⁰C

B. 4/9⁰C

C. 9/2⁰C

D. 9/4⁰C

B

## DETAILS…

If a piece of substance of specific heat capacity 450Jkg⁻¹k⁻¹ falls through a vertical distance of 20m from rest, the rise in temperature of the substance on hitting the ground when all its energies are converted into heat will be calculated as 4/9⁰C, let me explain…

The potential energy of the substance = mgh

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The heat energy of the same substance = mcΔT

Where;

m = mass of the substance,

g = acceleration due to gravity,

h = height of substance location from the ground,

c = specific heat capacity of the substance, and

ΔT = change in temperature.

If all the potential energy of the substance is converted to heat, we have:

mgh = mcΔT

mgh = mcΔT

gh = cΔT

therefore,

ΔT = gh/c

ΔT = (10 × 20)/450

ΔT = 200/450

ΔT = 4/9K = 4/9⁰C

### Now for the right answer to the above question:

1. Option A is incorrect.
2. Option B is incorrect.
3. C is incorrect.
4. D is the correct answer.

## KEY-POINTS…

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