# What is the final temperature of 1370 g of water, originally at 20.9 degrees C, if it absorbs...

## Question:

What is the final temperature of {eq}1370\ g {/eq} of water, originally at {eq}20.9 ^\circ C {/eq}, if it absorbs {eq}\rm 47.6\ kJ {/eq} of heat?

## Heat Transfer:

Heat can be transferred to substances and the result of which is a corresponding change in temperature. We can determine the specific heat of a substamce, {eq}\displaystyle c {/eq}, through heat transfer experiments using the equation, {eq}\displaystyle q = mc\Delta T {/eq}, where {eq}\displaystyle q {/eq} is the transferred heat, {eq}\displaystyle m {/eq} is the mass of the substance, and {eq}\displaystyle \Delta T {/eq} is the change in temperature.

## Answer and Explanation:

Determine the final temperature of water, {eq}\displaystyle T_f {/eq}, using the equation, {eq}\displaystyle q = mc(T_f - T_i) {/eq}, where {eq}\displaystyle q = 47.6\ kJ = 47600\ J {/eq} is the heat absorbed, {eq}\displaystyle m = 1370\ g {/eq} is the mass, {eq}\displaystyle c = 4.186\ \rm{J/g ^\circ C} {/eq} is the specific heat of water, anf {eq}\displaystyle T_i = 20.9 ^\circ C {/eq} 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{47600\ J}{1370\ g\times 4.186\ \rm{J/g ^\circ C}} + 20.9 ^\circ C&= T_f\\ 29.2 ^\circ C &= T_f \end{align} {/eq}

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from High School Physics: Help and Review

Chapter 17 / Lesson 12