The density of a 1.0 M solution of NaCl is 1.04 g/mol and its specific heat, C, is 3.60 J/C*g....

Question:

The density of a 1.0 M solution of NaCl is 1.04 g/mol and its specific heat, C, is 3.60 J/C{eq}\cdot {/eq}g. How much heat is absorbed by 31.4 mL of this solution if the temperature rises from 18.6 degrees C to 29.7 degrees C?

Heat Transfer:

When we transfer heat to a substance, it might change in temperature, if no chemical reaction or phase change occurs. The change in the temperature, {eq}\displaystyle \Delta T {/eq}, is proportional to the heat supplied, {eq}\displaystyle q {/eq}, such that it can be described by the equation, {eq}\displaystyle q = mc\Delta T {/eq}, where {eq}\displaystyle m {/eq} is the mass and {eq}\displaystyle c {/eq} is the specific heat of the substance.

Answer and Explanation:

Determine the total heat absorbed, {eq}\displaystyle q {/eq}, by the solution using the equation, {eq}\displaystyle q = mc\Delta T {/eq}, where {eq}\displaystyle m {/eq} is the mass of the solution, {eq}\displaystyle c = 3.60\ \rm{J/g ^\circ C} {/eq} is the specific heat, and {eq}\displaystyle \Delta T = 29.7 ^\circ C- 18.6 ^\circ C =11.1 ^\circ C {/eq} is the change in themperature. We determine the mass of the solution by multiplying the given volume, {eq}\displaystyle V = 31.4\ mL {/eq}, to the density, {eq}\displaystyle \rho = 1.04\ \rm{g/mol} {/eq}, or {eq}\displaystyle m = V\times \rho = 31.4\ mL\times 1.04\ \rm{g/mol} =32.656\ g {/eq}. We proceed with the solution.

{eq}\begin{align} \displaystyle q &= mc\Delta T\\ &= 32.656\ g\times 3.60\ \rm{J/g ^\circ C}\times 11.1 ^\circ C\\ &\approx 1300\ J \end{align} {/eq}


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Thermal Expansion & Heat Transfer

from High School Physics: Help and Review

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