Heat is supplied uniformly at the rate of 100W to 1.0 × 10¯²kg of a liquid for 20 seconds. If the temperature of the liquid rises by 5°C, then the specific heat capacity of the liquid is

A. 2.0 × 10²Jkg¯¹K¯¹

B. 2.0 × 10²Jkg¯¹

C. 4.0 × 10⁴ Jkg¯¹K¯¹

D. 4.0 × 10⁴ Jkg¯¹

QUICK ANSWER…

C

DETAILS…  

First of all, you should quickly strike out options B and D.

This is because specific heat capacity is the quantity of heat energy needed to change the temperature of 1kg of a substance by 1 kelvin. Therefore, the right unit for specific heat capacity is Jkg¯¹K¯¹. let’s solve for the specific answer…

Data given:

Power = 100w

Mass = 1.0 × 10¯²kg

Time = 20s

Change in temperature = 5°c = 5K

FORMULA:

H = mcθ

Where H = heat supplied,

m = mass of the liquid,

c = specific heat capacity,

and θ = change in temperature

but if this heat is coming from an electric set up, with power consumed, then:

H = power × time = 100 × 20 = 2000J

Hence, substituting these into the first formula, we have:

H = mcθ

2000 = 1.0 × 10¯² × c × 5

2000 = 5 × 10¯² × c

C = 2000/(5 × 10¯²)

C = (2000/5) × 10²

C = 400 × 10²

C = 4 × 10⁴

C = 4 × 10⁴ Jkg¯¹K¯¹

Now for the right answer to the above question:

1. A is not the right answer.
2. B is incorrect. wrong unit.
3. C is correct.
4. D is incorrect. wrong unit.

KEY-POINTS…

You may please note this/these:

• The change in temperature did not change even as the unit changed from °c to K. this is because we are dealing with a net change and not a specific temperature.
• The unit of specific heat capacity is Jkg¯¹K¯¹ and not Jkg¯¹

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/ culled from 2020 JAMB-UTME physics past question 32 /