In a constant-pressure calorimetry experiment, a reaction gives off 20.8 kJ of heat. The...

Question:

In a constant-pressure calorimetry experiment, a reaction gives off {eq}20.8\ kJ {/eq} of heat. The calorimeter contains {eq}150\ g {/eq} of water, initially at {eq}21.4 ^\circ {/eq} Celsius. What is the final temperature of the water? The heat capacity of the calorimeter is negligibly small.

Heat Transfer:

The heat transferred to a substance is a function of the change in temperature, {eq}\displaystyle \Delta T {/eq}, mass, {eq}\displaystyle m {/eq}, and specific heat, {eq}\displaystyle c {/eq}, of the substance, such that {eq}\displaystyle q = mc\Delta T {/eq}. Values involved in this equation are typically involved in calorimetry experiments, such that we can relate the heat transferred to a heat of a certain reaction.

Determine the final temperature, {eq}\displaystyle T_f {/eq}, in the given system by neglecting the absorbed heat by the calorimeter and considering only the heat absorbed by the water using the equation, {eq}\displaystyle q = mc(T_f-T_i) {/eq}, where q = 20.8 kJ = 20800 J is the absorbed heat, c = 4.186 J/g{eq}\displaystyle ^\circ C {/eq}, and T{eq}\displaystyle _i {/eq} = 21.4{eq}\displaystyle ^\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{20800\ J}{150\ g\times 4.186\ J/g^\circ C}+ 21.4 ^\circ C &= T_f\\ 54.5^\circ C &\approx T_f \end{align} {/eq}