Physics, asked by Anonymous, 2 days ago

An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10 ℃ to 15℃ in 100 s.
Calculate :
(i)the specific heat capacity of the liquid.
(ii)the heat capacity of 4.0 kg of liquid.

Answers

Answered by 24Karat
4

\huge\mathcal{\fcolorbox{lime}{black}{\blue{Question :-}}}

An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10 ℃ to 15℃ in 100 s.

Calculate :

(i)the specific heat capacity of the liquid.

(ii)the heat capacity of 4.0 kg of liquid.

\huge\mathcal{\fcolorbox{lime}{black}{\blue{AnsweR :-}}}

Power of heater P = 600 W

Mass of liquid m = 4.0 kg

Change in temperature of liquid = (15 - 10)℃ = 5 ℃ (or 5 K)

Time taken to raise its temperature = 100s

Heat energy required to heat the liquid =

ΔQ = mcΔT

and,

ΔQ = P × t

c = ΔQ/mΔT = 60000/4 × 5 = 3000Jkg^-1K^-1

Heat capacity= c x m

Heat capacity= 4 x 3000JKg^-1K^-1 = 1.2 x 104 J/K

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Answered by Anonymous
0

Answer:

The degree symbol or degree sign, °, is a typographical symbol that is used, among other things, to represent degrees of arc, hours, degrees of temperature or alcohol proof. The symbol consists of a small superscript circle

Explanation:

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