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Difficulty: MediumExothermic and Endothermic Reactions

When a 2.0 g2.0\text{ g} sample of a solid solute is completely dissolved in 100.0 g100.0\text{ g} of distilled water initially at 25.0∘C25.0^\circ\text{C}, an exothermic process occurs and the temperature of the water rises to 30.0∘C30.0^\circ\text{C}. Assuming the specific heat capacity of water is 4.2 J g−1 K−14.2\text{ J g}^{-1}\text{ K}^{-1} and neglecting the heat capacity of the calorimeter container, what is the amount of heat energy, in Joules, absorbed by the water?

Answer: 2100 J

Answer

The amount of heat energy absorbed by the water is 2100 J2100\text{ J}.
In an exothermic process, heat is transferred to the surroundings (water). Using the calorimetric relation Q=mcΔTQ = m c \Delta T with m=100.0 gm = 100.0\text{ g}, c=4.2 J g−1 K−1c = 4.2\text{ J g}^{-1}\text{ K}^{-1}, and ΔT=(30.0−25.0) K=5.0 K\Delta T = (30.0 - 25.0)\text{ K} = 5.0\text{ K}, the heat gained by the water is Q=100.0×4.2×5.0=2100 JQ = 100.0 \times 4.2 \times 5.0 = 2100\text{ J}.

Step-by-Step Solution

1
Determine the change in temperature (ΔT\Delta T) of the water.
ΔT=Tfinal−Tinitial=30.0∘C−25.0∘C=5.0∘C=5.0 K\Delta T = T_{\text{final}} - T_{\text{initial}} = 30.0^\circ\text{C} - 25.0^\circ\text{C} = 5.0^\circ\text{C} = 5.0\text{ K}
The heat calculation requires the temperature difference resulting from the exothermic dissolution.
2
Apply the heat formula Q=mcΔTQ = m c \Delta T.
Q=100.0 g×4.2 J g−1 K−1×5.0 KQ = 100.0\text{ g} \times 4.2\text{ J g}^{-1}\text{ K}^{-1} \times 5.0\text{ K}
Heat absorbed depends directly on the mass of water, its specific heat capacity, and the temperature rise.
3
Calculate the total heat energy absorbed.
Q=2100 JQ = 2100\text{ J}
Multiplying the mass, heat capacity, and temperature change yields 2100 J2100\text{ J}.

Key Concept

Calorimetric Heat Calculation (Q=mcΔTQ = m c \Delta T)
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Topics

Exothermic and Endothermic Reactions