Question

1. Calculate the heat lost by the hot water.

2. Calculate the heat gained by the cold water.

3. Calculate the heat gained by the calorimeter.

4. Calculate the calorimeter constant.

Data:

Atmospheric Pressure- 100.87 kPa

Boiling point of water at this pressure- 99.8°C

Mass of empty calorimeter- 8.3676 grams

Mass of calorimeter plus 40 mL of water- 48.6923 grams

Mass of calorimeter after adding 50 mL of hot water- 98.0557 grams

Temperature of cold water just before adding addition of hot water- 21.7°C

Tmax- 61.3°C

Delta Tc- 39.7°C

Delta Thw- -37.5°C

(2 styrofoam cups were used for the calorimeter and there was no lid on it so it was open to the air)

Answer 1

Ans. #1. Hot water:

Mass of hot water = Mass of (hot water + Calorimeter + Cold water) - Mass of (Calorimeter + Cold water)

                                    = 98.0557 g – 48.6923 g = 49.3634 g

dT for hot water = -37.5⁰C (the -ve sign indicates cooling of hot water by heat loss)

# Heat lost by hot water, q_hw (putting the values in equation 1) –

            q_hw = 49.3634 g x 4.184 H g^-10C^-1 x (-37.5⁰C) = 7745.11746 J

#2. Cold water:

Mass of cold water = Mass of (Calorimeter + Cold water) - Mass of Calorimeter

                                    = 48.6923 g – 8.3676 g = 40.3247 g

dT for cold water = 39.7⁰C (given, dTc)

# Heat gained by cold water, q_cw (putting the values in equation 1) –

            q_cw = 40.3247 g x 4.184 H g^-10C^-1 x 39.7⁰C = 6698.12622856 J

#3. Heat gained by calorimeter, q_cal = q_hw - q_cw = 7745.11746 J - 6698.12622856 J

                                                            = 1046.99123144 J

#4. Calorimeter constant = Heat gained by calorimeter / dT for calorimeter

                                                = 1046.99123144 J / 39.7⁰C

                                                = 36.3726 J ⁰C^-1

Note: It’s assumed that dT for calorimeter is equal to that of cold water.