A piece of substance of specific heat capacity 450Jkg⁻¹k⁻¹ falls…
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A piece of substance of specific heat capacity 450Jkg⁻¹k⁻¹ falls through a vertical distance of 20m from rest. Calculate the rise in temperature of the substance on hitting the ground when all its energies are converted into heat. [g = 10ms⁻²]
A. 2/9⁰C
B. 4/9⁰C
C. 9/2⁰C
D. 9/4⁰C
QUICK ANSWER…
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
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:
- Option A is incorrect.
- Option B is incorrect.
- C is incorrect.
- D is the correct answer.
KEY-POINTS…
You may please note these/this:
- Even though this cannot happen in real life, the assumption that all the potential energy got converted into heat is a mere theoretical exercise.
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/ culled from 2021 JAMB-UTME PHYSICS question 29 /
