Question

If a reaction produces 1.506 kJ of heat, which is trapped in 30.0 g of water initially at 26.5 °C in a calorimeter like that in Figure 5.12, what is the resulting temperature of the water?

Short answer

The resulting final temperature of the water is 38.5 °C.

Step-by-step solution

Rise in temperature

Given information:

The mass of the water (m) = 30.0 g

The specific heat of the water (C) = 4.18 J/g°C

The heat evolved (Q) = 1.506kJ or 1506J

Initial temperature (T) = 26.5°C

Calculation of temperature

The heat evolved during the reaction is mathematically presented as\({\bf{Q = mC\Delta T}}\). Therefore, by putting all the given values in \({\bf{Q = mC\Delta T}}\), we will get a rise in the temperature.

\({\rm{1506 = 30 }} \times {\rm{ 4}}{\rm{.18(}}{{\rm{T}}_{\rm{f}}}{\rm{ - 26}}{\rm{.5)}}\)

\({\rm{1506 = 125}}{\rm{.4(}}{{\rm{T}}_{\rm{f}}}{\rm{ - 26}}{\rm{.5)}}\)

\({\rm{1506 = 125}}{\rm{.4}}{{\rm{T}}_{\rm{f}}}{\rm{ - 3323}}{\rm{.1}}\)

\({\rm{125}}{\rm{.4}}{{\rm{T}}_{\rm{f}}}{\rm{ = 4829}}{\rm{.1}}\)

\({{\rm{T}}_{\rm{f}}}{\rm{ = 38}}{\rm{.5}}^\circ {\rm{C}}\)