Question

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

Answer 1

Answer:

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,

$Q=mc\Delta T$

Where

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

So,

$6000=200\times 4.18\times (T_2-298)\\\\\dfrac{6000}{200\times 4.18}=(T_2-298)\\\\7.17=(T_2-298)\\\\7.17+298=T_2\\\\T_2=305.17\ K$

So, the final temperature is equal to 305.17  K.