50 ml of water at 46.9 degrees C were mixed with 50 ml of water at 25.1 degrees C in a...


{eq}50\ ml {/eq} of water at {eq}46.9 ^\circ C {/eq} were mixed with {eq}50\ ml {/eq} of water at {eq}25.1 ^\circ C {/eq} in a calorimeter also at {eq}25.1 ^\circ C {/eq}. The final temperature was {eq}30.1 ^\circ C {/eq}. Assuming that neither the density of water nor its specific heat capacity change with temperature, calculate the total heat capacity of the calorimeter. Density of water = {eq}1.00\ g/mL {/eq}, Specific heat capacity = {eq}4.18\ J / g \cdot K {/eq}.

Heat Transfer:

In calorimetry, we typically allow substances to transfer heat among each other. This means that the heat gained by one substance, {eq}\displaystyle q_{gained} {/eq}, is equated to the heat lost of another, {eq}\displaystyle -q_{lost} {/eq}. With this, we use the equation, {eq}\displaystyle q_{gained } = -q_{lost} {/eq}, in order to solve problems on heat transfer.

Answer and Explanation:

Determine the heat capacity of the calorimeter, {eq}\displaystyle c_{cal} {/eq}, by equating the heat lost by the hot water, {eq}\displaystyle -q_{hot} {/eq}, by the heat gained by the cooler water, {eq}\displaystyle q_{cool} {/eq}, and the calorimeter, {eq}\displaystyle q_{cal} {/eq}. We follow the general equation, {eq}\displaystyle -q_{hot} = q_{cool}+q_{cal} {/eq}. We use the heat transfer equation for the heat released or gained by the water, {eq}\displaystyle q = mc\Delta T {/eq}. Moreover, we must use the identity, {eq}\displaystyle q_{cal} = c_{cal}\Delta T {/eq}. We use the following values for the variables:

  • {eq}\displaystyle m_{hot} = 50\ mL\times 1\ g/mL = 50\ g {/eq}
  • {eq}\displaystyle c = 4.18\ J/g\ K = 4.18\ J/g ^\circ C {/eq}
  • {eq}\displaystyle \Delta T_{hot} = 30.1 ^\circ C - 46.9 ^\circ C = -16.8 ^\circ C {/eq}
  • {eq}\displaystyle m_{cool} = 50\ mL\times 1g/ mL = 50\ g {/eq}
  • {eq}\displaystyle \Delta T_{cool} = 30.1 ^\circ C - 25.1 ^\circ C = 5.0 ^\circ C {/eq}
  • {eq}\displaystyle \Delta T_{cal} =30.1 ^\circ C - 25.1 ^\circ C = 5.0 ^\circ C {/eq}

We proceed with the solution.

{eq}\begin{align} \displaystyle -q_{hot} &= q_{cool}+q_{cal}\\ -m_{hot} c\Delta T_{hot} &= m_{cool}c\Delta T_{cool} + c_{cal}\Delta T_{cal}\\ -m_{hot} c\Delta T_{hot}- m_{cool}c\Delta T_{cool}&= c_{cal}\Delta T_{cal}\\ \frac{-m_{hot} c\Delta T_{hot}- m_{cool}c\Delta T_{cool}}{\Delta T_{cal}}&= c_{cal}\\ \frac{-50\ g\times 4.18\ J/g ^\circ C\times -16.8 ^\circ C - 50\ g\times 4.18J/g ^\circ C\times 5.0 ^\circ C }{5.0 ^\circ C} &= c_{cal}\\ \boxed{\rm 493\ J/^\circ C } &\approx c_{cal} \end{align} {/eq}

Learn more about this topic:

Thermal Expansion & Heat Transfer

from High School Physics: Help and Review

Chapter 17 / Lesson 12

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