Question

assume you are performing the calibration step of experiment 8 and you begin with 80 g of water at 20 oc and 80 g of water at 80 oc. after adding the two portions of water into your calorimeter setup and following the procedure outlined in the experiment, you determine the temperature of the mixed portions of water to be 45 oc. what is the heat capacity of the calorimeter?

Answer 1

The heat capacity of the calorimeter is 133.76 J°/C.

A calorimeter is an apparatus employed when estimating the amount of heat required in a chemical reaction.

For the given calorimetric function:

• the heat lost by the hot water = to the heat gained by the cold water and the calorimeter.

These can be computed mathematically as:

$\mathbf{m_wC_w\Delta \theta _w = m_cC_c\Delta \theta _c + C_{cal} \Delta_c}$

$\mathbf{80 \ g\times 4.18 J/g^0 C \times (80-45)^0 \ C = 80 g \times 4.18\ J/g^0 \ \times (45- 20)^0 +C_{cal} \times (45- 20)^0 \ C}$

$\mathbf{11704 \ J =8360 \ J+C_{cal} \times (25)^0 \ C}$

$\mathbf{(11704 -8360) \ J= C_{cal} \times (25)^0 \ C}$

$\mathbf{ C_{cal} = \dfrac{3344 \ J} { (25)^0 \ C}}$

$\mathbf{ C_{cal} =133.76 \ J/^0 C}$

Therefore, the heat capacity of the calorimeter is 133.76 J°/C.

Learn more about calorimeter here:

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