# A 45 kg sample of water absorbs 349 kJ of heat. If the water was initially at 25.5^oC, what is...

## Question:

A 45 kg sample of water absorbs 349 kJ of heat. If the water was initially at {eq}25.5^oC {/eq}, what is its final temperature?

## Heat Transfer:

Heat transferred to a substance will result in the change in its temperature. We relate the heat, q, that is transferred or absorbed by the substance to the change in temperature, {eq}\displaystyle \Delta T {/eq}, using the following equation:

{eq}\displaystyle q = mc\Delta T {/eq}

In this equation, m is the mass of the sample of substance and c is its specific heat.

We can determine the final temperature of water, {eq}\displaystyle T_f {/eq}, by applying the heat transfer equation:

{eq}\displaystyle q= mc(T_f-T_i) {/eq}, where:

• q = 349 kJ = 349,000 J is the absorbed heat
• m = 45 kg = 45000 g is the mass
• c = 4.186 J/g{eq}\displaystyle ^\circ {/eq}C is the specific heat
• {eq}\displaystyle T_i {/eq} = 25.5{eq}\displaystyle ^\circ {/eq}C is the initial temperature

We proceed with the solution:

{eq}\begin{align} \displaystyle q &= mc(T_f-T_i)\\ \frac{q}{mc} &= T_f-T_i\\ \frac{q}{mc}+T_i &= T_f\\ \frac{349000\ J}{45000\ g\times 4.186\ J/g^\circ C} + 25.5^\circ C &= T_f\\ \boxed{37.4^\circ C}&\approx T_f\\ \end{align} {/eq}