Question

The specific heat of water is 4.184Jg ∘C. Determine the final temperature when 600.0 g water at 75.5∘C absorbs 5.90×104 J of energy.]

Answer 1

Answer:

$T_2=98.5^{\circ}$

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 :

$Q=mc\Delta T\\\\Q=mc(T_2-T_1)\\\\\dfrac{Q}{mc}=(T_2-T_1)\\\\\\T_2=\dfrac{Q}{mc}+T_1\\\\T_2=\dfrac{5.9\times 10^4}{600\times 4.184}+75\\\\T_2=98.5^{\circ}$

So, the final temperature is equal to $98.5^{\circ}$.