If 6000 J of heat is added to 200 gm of water at 25° C. What will be its final
temperature?​

T₂ = 305.17 K

Explanation:

Given that,

Heat, Q = 6000 J

Mass, m = 200 gram

Initial temperature, T₁ = 25° C

We need to find its final temperature. Let it is T₂.

We know that,

Where

c is the specific heat of water, c = 4.18 J/g°C

So,

So, the final temperature is equal to 305.17  K.

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