Question

to raise the temperature from 27°C to 77°C of 500g substance 200J capacity.calculate the specific heat capacity.​

Answer 1

Answer:-

Given:

Heat absorbed (Q) = 200 J

Original temperature = 27° C

New temperature = 77° C

Mass of the substance (m) = 500 g.

We know that,

Q = ms∆T

(∆T is difference between the temperatures)

⟹ 200 = 500/1000 * s * (77 - 27)

• 1 g = 1/1000 kg
• 500 g = 500/1000

⟹ 200 * 100/500 * (1/50 + 273°) = s

• °K = °C + 273°

⟹ 0.12 J * kg/K = s

Therefore, the specific heat of the substance is 0.12 J * kg/K.