Question

100 points!
Physics (HEAT )

Answer with 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

Answer:

Water equivalent refers to the amount of water taking up the same amount of energy as that of the paraffin oil (in this case) with the same rise in temperature.

Calculation:

Let mass of water be "m" , specific heat capacity be "c" and change in temperature be "∆θ"

So, let heat transferred be Q.

∴ Q = m × c × ∆θ

=> 4180 × 10³ = m × 4.18 × (320 - 300)

=> 4180 × 10³ = m × 4.18 × 20

=> m × 20 = (4.18/4.18) × 10^6

=> m = 1/2 × 10^5

=> m = 50 × 10³ g

=> m = 50 kg

So the final answer is 50 kg.

Answer 2

$\huge{\orange{\underline{\red{\mathtt{GIVEN:-}}}}}$

SPECIFIC HEAT OF WATER 4.18 Jg^-1 ℃^-1

calory =4.18j.

temperacture =300k to 320k. (320-300)=20k.

$\huge{\orange{\underline{\red{\mathtt{FORMULA:-}}}}}$

Q=m×c×Δ$\theta$

$\huge{\orange{\underline{\red{\mathtt{ANSWER:-}}}}}$

$\rightarrow$ 4180×10³=m×4.18×20

$\rightarrow$M×20=(4.18/4.18)×10^6

$\rightarrow$M=½×10^5

$\rightarrow$M=50×10³g.

[NOTE :- 10³g convert into kg]

.°.$\rightarrow$M=50kg.

$\huge{\orange{\underline{\red{\mathtt{THANKS.}}}}}$