Question

A student is preparing to perform a series of calorimetry experiments. She first wishes to determine the calorimeter constant (Ccal) for her coffee cup calorimeter. She pours a 50.0 mL sample of water at 345 K into the calorimeter containing a 50.0 mL sample of water at 298 K. She carefully records the final temperature of the water as 317 K. What is the value of Ccal for the calorimeter

Answer 1

Explanation:

The given data is as follows.

     $V_{1}$ = 50 ml,      $T_{1}$ = 345 K

     $T_{2}$ = 298 K,      $T_{f}$ = 317 K,

    $V_{2}$ = 50 ml

Now, we will calculate the heat energy as follows.

        $Q_{hot} = m \times C_{p} \times (T_{1} - T_{f})$

                     = $50 g \times 4.184 \times (345 - 317)$

                     = 5857.6 J

       $Q_{cold} = m \times C_{p} \times (T_{f} - T_{2})$

                     = $50 g \times 4.184 \times (317 - 298)$

                     = 3974.8 J

As,   $Q_{hot} = -Q_{cold}$ so there will be loss of heat. And, some heat will go to the calorimeter.

Hence,     $Q_{hot} = Q_{cold} + Q_{cal}$

                 $5857.6 = 3974.8 + Q_{cal}$

               $Q_{cal}$ = 1882.8 J

We assume that the temperature of (calorimeter + water) is 298 K. Hence,

                  dT = $T_{f} - T_{2}$

                        = (317 - 298) K

                        = 19 K

Hence, we will calculate the specific heat as follows.

               C = $\frac{Q}{dT}$

                   = $\frac{1882.8 J}{19}$

                   = 99.1 J/K

Thus, we can conclude that the value of $C_{cal}$ for the calorimeter is 99.1 J/K.