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,
[tex]Q=mc\Delta T[/tex]
Where
c is the specific heat of water, c = 4.18 J/g°C
So,
[tex]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[/tex]
So, the final temperature is equal to 305.17 K.
