Question

100 points!
Physics (HEAT )

[NOTE - DON'T KNOW THEN DON'T ANSWER ]

Answer with CORRECT STEPS

Find the water equivalent of paraffin oil if 100 kg of paraffin oil absorbs 4180 × 10 ^ 3 J to raise its temperature from 300 K to 320 K.

[Specific heat of water = 4.18 Jg^-1 ℃^-1]
Ans :- 50 kg ​

Answer 1

Explanation:

Given that,

Heat absorbed(Q) = 4180 × 10 ³ J

Specific heat(C) of water = 4.18 Jg^-1 ℃^-1

T1 = 300 K

T2 = 320 K

let the mass of water be m

Q = mCΔT

4180 × 10³ = m × 4.18 × (320 - 300)

4180 × 10³ = m × 4.18 × 20

m = 50 Kg

Answer 2

Explanation:

According to question,

mcΔT = 4180 × 10³

m is mass of water

c is Specific heat of water = 4.18 Jg^-1 ℃^-1

ΔT = temperature difference

putting the value,

m × 4.18 Jg^-1 ℃^-1 (320 - 300) = 4180 × 10³

m = 50 Kg