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Chapter 5: Q23E (page 270)

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

Short Answer

Expert verified

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

Step by step solution

01

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掳C

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

Initial temperature (T) = 26.5掳C

02

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.

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

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

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

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

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

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