Answer:
[tex]T_2=98.5^{\circ}[/tex]
Explanation:
Given that,
The specific heat of water is 4.184Jg°C
Mass, m = 600 g
Initial temperature, T₁ = 75.5°C
We need to find the final temperature. We know that heat absorbed is given by :
[tex]Q=mc\Delta T\\\\Q=mc(T_2-T_1)\\\\\dfrac{Q}{mc}=(T_2-T_1)\\\\\\T_2=\dfrac{Q}{mc}+T_1\\\\T_2=\dfrac{5.9\times 10^4}{600\times 4.184}+75\\\\T_2=98.5^{\circ}[/tex]
So, the final temperature is equal to [tex]98.5^{\circ}[/tex].
