Question

PART A: A student constructs a coffee cup calorimeter and places 50.0 mL of water into it. After a brief period of stabilization, the temperature of the water in calorimeter is determined to be 19.6 °C. To this is added 50.0 mL of water that was originally a temperature of 54.5 °C. A careful plot of the recorded temperature established T0 as 31.1 °C. What is the calorimeter constant (J/°C)?

DensityH2O = 1.00 g/mL

Specific HeatH2O = 4.184 J/g·°C

PART B: You placed 43.3 g of an unknown metal at 100 °C into a coffee cup calorimeter that contained 50.0 g of water that was initially at 22.0 °C. The equilibrium temperature of mixing (T0) was determined to be 23.1 °C. The calorimeter constant was known to be 51.5 J/°C.

Specific HeatH2O = 4.184 J/g·°C

-What is the total amount of heat (J) lost by the metal?

-What was the specific heat (J/g·°C) of the metal?

(need both heats)

Answer 1

Part A

Density of water=1 g/mL

Mass of water in coffee cup calorimeter=Volume x Density=50.0 mL x 1.00 g/mL=50.0 g

Initial temperature of water in calorimeter=19.6°C

Mass of water added=Volume x Density=50.0 mL x 1.00 g/mL

=50.0 g

Initial temperature of water added=54.5°C

Final temperature of the system=31.1°C

Heat gained by water in calorimeter=mass of water x specific heat x rise in temperature

=50.0 g x 4.184 J/g°C x (31.1°C -19.6°C)

=50.0 g x 4.184 J/g°C x 11.5°C=2405.8 J

Heat lost by water added=mass of water x specific heat x(fall in temperature)

=50.0 g x 4.184 J/g°C x (54.5°C - 31.1°C)

=50.0 g x 4.185 J/g°C x 23.4 °C

=4895.28 J

The remaining heat=4895.28 J -2405.8 J=2489.48 J is absorbed by the calorimeter

Calorimeter constant=Heat absorbed by calorimeter/rise in temperature of water in calorimeter

=2489.48 J/11.5°C=216.48 J/°C

Part B

Mass of metal=43.3 g

Initial temperature of metal=100°C

Mass of water=50.0 g

Initial temperature of water=22.0 °C

Equilibrium temperature of the system=23.1 °C

Specific heat of water=4.184 J/g°C

Let specific heat of the metal be c

Heat lost by metal=Heat gained by water

Mass of metal x specific heat of the metal x fall in temperature of metal=Mass of water x specific heat of the water x rise in temperature of water+calorimeter constant x rise in temperature of water in calorimeter

43.3 g x c x (100°C-23.1°C)= 50.0 g x 4.184 J/g°C x (23.1 °C-22.0 °C)+51.5 J/°C x (23.1 °C-22.0°C)

43.3 g x c x 76.9 °C=50.0 g x 4.184 J/g°C x 1.1 °C+51.5 °C x 1.1 °C

c=(50.0 g x 4.184 J/g°C x 1.1 °C + 51.5 J/°C x 1.1 °C)/(43.3 g x 76.9°C)=(230.12 J+56.65 J)/3329.77 g°C=286.77 J/3329.77 g°C

=0.086 J/g°C

Total amount of heat (J) lost by the metal=mass of metal x specific heat of the metal x fall in temperature

=43.3 g x 0.086 J/g°C x 76.9 °C=286.36 J