Question

❥❥ Hey Mate!

❥❥An electric heater of power 1000 W raises the temperature of 5 kg of a liquid from 25°C to 31°C in 2 minutes. Calculate:-

1. The heat Capacity
2. The specific heat Capacity of liquid.

Spamming = 10 relevant answers reported. ​

Answer 1

Answer:

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Explanation:

Heat is: Q=mc∆t => Pxt=mc∆t => 1000x2=5xcx6 => c=200/3 J/K

Answer 2

❥ Answer :-

Given,

Time t = 2 minutes

= 2 × 60s = 120s

Rise in temperature T = ( 31 - 25 ) ° C = 6°C = 6K

Mass of liquid m = 5 kg

Energy supplied by the heater = power × time

Q = 1000 W × 120s = 1.2 × 10⁵ J

i. Heat Capacity

C' = energy supplied Q/ rise in temperature T

= 1.2 × 10⁵ J/ 6 K = 2 × 10⁴ J K‐¹

ii. Specific Heat Capacity

c = heat capacity (C')/ mass (m)

= 2 × 10⁴ J K ‐¹/ 5 kg

= 4 × 10³ J kg-¹ K-¹

❥ Hope it helps u Anshu