Question

A neutralization reaction was carried out in a calorimeter. The change in temperature (∆T) of the solution was 10.9 °C and the mass of the solution was 100.0 g. Calculate the amount of heat energy gained by the solution (qsol). Use 4.18 J/(g•°C) as the specific heat, Cs, of the solution

Answer 1

Answer:

The correct answer is 4,556.2 J

Explanation:

The amount of heat is calculated with the following equation:

Heat= m x Cs x ΔT

Given:

m= 100.0 g (mass)

Cs= 4.18 J/g.ºC (specific heat)

ΔT= 10.9ºC (change in temperature)

We introduce the data in the equation and calculate the heat:

Heat = 100.0 g x 4.18 J/ºC.g x 10.9ºC = 4,556.2 J

Therefore, the amount of heat absorbed by the water (gained by the solution) is 4,556.2 J

Answer 2

The amount of heat gained by the solution in the neutralization reaction has been 4.5562 kJ.

The specific heat can be defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius. The expression of specific heat can be given as:

$Q=mc\Delta T$

Where, heat has been, $Q$.

The mass of the substance, $m=100\;\rm g$

The specific heat of solution, $c=4.18\;\rm J/g/^\circ C$

The change in temperature, $\Delta T=109^\circ \rm C$

Substituting the values:

$Q=\rm 100\;g\;\times\;4.18\;J/g/^\circ C\;\times\;100\;^\circ C\\\textit Q=4,556.2\;J\\\textit Q=4.5562\;kJ$

The amount of heat gained by the solution in the neutralization reaction has been 4.5562 kJ.

For more information about specific heat, refer to the link:

https://brainly.com/question/15177263