Question

Info for part 1: Determining the heat capacity of the calorimeter

Mass of the empty Styrofoam cups and cover: 30.21 g

Mass of the cups, cover and 70 ml cool water: 95.21 g

Mass of the cups, cover, 70 ml water and 30 ml hot water: 127.41 g

Initial temperature of the hot water (boiling bath): 99.5 C

Initial temperature of the cool water in the cups: 18 C

Final temperature of the calorimeter and the added hot water: 43 C

Calulations: Please show work I am so confused

a)Mass of the hot water (m_hot) (g)

b)Temperature change of the hot water (J)

c)Heat lost by the hot water (J)

d)Mass of the cool water (m_cool) J/C

e)Temperature change of the cool water J/C

f)Heat absorbed by the cool water

g) Heat gained by the calorimeter

h)Heat capacity of the calorimeter

i)Average Heat capacity of the calorimeter from both trials

J)Initial temperature of the hot metal (boiling bath): 30.21 g

K) Initial temperature of the cool water in the cups: 131.07 g

L)Final temperature of the calorimeter and the added hot metal: 84.67 g

Part 2 info: determining the heat capacity of a metal

Mass of the empty Styrofoam cups and cover: 30.21 g

Mass of the cups, cover and 100 ml water:131.07 g

Mass of the metal (m_m):84.67 g

Initial temperature of the hot metal (boiling bath): 99.3 C

Initial temperature of the cool water in the cups: 21.3 C

Final temperature of the calorimeter and the added hot metal: 27.4 C Calculations :( (metal is nickel) -Mass of the cool water (mcool) -Temperature change of the cool water -Heat absorbed by the cool water -Average heat capacity of the calorimeter (Hcal) -Heat gained by the calorimeter -Heat lost by metal -Temperature change of the hot metal -Specific Heat of metal -Average Specific Heat of metal -percent error

Answer 1

1a) Mass of hot water (m_hot) = (mass of cup, cover, 70 mL cold water and 30 mL hot water) – (mass of cup, cover and 70 mL cold water) = (127.41 g) – (95.21 g) = 32.2 g.

b) Temperature change of hot water = (initial temperature of hot water) – (final temperature of the calorimeter and added hot water) = (99.5 °C) – (43 °C) = 56.5 °C.

c) Heat lost by the hot water = (mass of hot water)*(specific heat capacity of water)*(temperature change of hot water) = (32.2 g)*(4.18 J/g.°C)*(56.5 °C) = 7604.674 J (specific heat capacity of water = 4.18 J/g.°C).

d) Mass of cold water (m_cold) = (mass of cup, cover and 70 mL cold water) – (mass of empty cup and cover) = (95.21 g) – (30.21 g) = 65.00 g.

e) Temperature change of the cold water = (final temperature of the calorimeter and the added hot water) – (initial temperature of the cold water) = (43 °C) – (18 °C) = 25 °C.

f) Heat absorbed by the cold water = (mass of cold water)*(specific heat of water)*(change in temperature of cold water) = (65.00 g)*(4.18 J/g.°C)*(25 °C) = 6792.500 J.

g) Heat gained by the calorimeter = (heat lost by the hot water) – (heat gained by the cold water) = (7604.674 J) – (6792.500 J) = 812.174 J.

h) Heat capacity of the calorimeter = (heat gained by the calorimeter)/(change in temperature of the cold water) = (812.174 J)/(25 °C) = 32.48696 J/°C.

We will assume that the calorimeter was initially at the temperature of the cold water and the temperature increased to 43°C.

i) You have furnished the data for a single trial; so I cannot take an average of two trials.