Question

Lab Experiment: Determining the Calorimeter Constant

NEED HELP WITH THE FOLLOWING BOLD TEXTS, PLEASE ANSWER ALL AND NOT PARTIAL. THANK YOU

Trial 1                  Trial 2

A. Volume of room temperature water, mL                                                     75mL                   100mL

B. Volume of hot water, mL                                                                             75mL                   100mL

C.Initial temperature of room temperature water, °C   23.3°C 24.5°C

D. Initial temperature of hot water, °C                                                             20.2°C 22.2°C

E. Final (max.) temperature of mixture, °C                                                     19.5°C 21.8°C

F. ΔT of room temperature water, °C           ________           ________

G. Mass of room temperature water, g    ________           ________

(assume a density of 1.0 g/mL)

H. q_rt, J ________           ________

In the space below, show your work for calculating q_rt for trial 1

I. ΔT of hot water, °C                                                                                       _______             ________

J. Mass of hot water, g _______             ________

K. q_hot, J    ________           ________

In the space below, show your work for calculatimg q_rt for trial 1

L. Heat absorbed by calorimeter, J     ________           ________

M. C_cal, J/°C                                                                                                        ________           ________

In the space below, show your work for calculating C_cal for trial 1.

N. Average C_cal                                                                                                                 ________

Answer 1

		Trial 1	Trial 2
A	Volume of room temperature water, mL	75	100
B	Volume of hot water, mL	75	100
C	Initial temperature of hot water, $^\circ C$	23.3	24.5
D	Initial temperature of room temp water, $^\circ C$	20.2	22.2
E	final temeprature of mixture, $^\circ C$	19.5	21.8
F	$\Delta T$ of hot water, $^\circ C$	3.8	2.7
G	Mass of hot water, g	75	100
H	hot , heat lost by hot water, J	1193.01	1130.22
I	$\Delta T$ of room temp water, $^\circ C$	0.7	0.4
J	Mass of room temp water, g	75	100
K	$q_{rt}$, heat gained by room temp water, J	209.765	167.44
L	Hear absorbed by calorimeter, J	983.245	962.78
M	CCal, J/C	258.74	356.58
N	Average Ccal, J/ C	307.66

note: In your data, temp of hot water is lower than room temp water, which is certainly a mistake. Hence I have duly exchanged the data for room temp water and hot water.

F.

Change in temperature hot water,

For Trial 1

$\Delta T_1 = |T(after \ mixing) - T(room \ temp \ water)| = |19.5 - 23.3| =3.8 ^\circ C$

For Trial 2

$\Delta T_2 = |T(after \ mixing) - T(room \ temp \ water)| =| 21.8 - 24.5| =2.7 ^\circ C$

G.

Mass of hot water.

For trial 1

Volume of room temp water = 75 mL

Density = 1g/mL

hence, mass = Volume* Density = 75 mL * 1 g/ mL = 75 g

For Trial 2

Volume of room temp water = 100 mL

Density = 1g/mL

hence, mass = Volume* Density = 100 mL * 1 g/ mL = 100 g

H.

Heat lost by hot water.

For trial 1

Heat gained by the room temp water is given by the formula

$q_{rt}= m \times C \times \Delta T = 75 \ g \times 4.186 \ J \ g^{-1} \ ^\circ C^{-1} \times 3.8 ^\circ C = 1193.01 \ J$

For Trial 2

$q_{rt}= m \times C \times \Delta T = 100 \ g \times 4.186 \ J \ g^{-1} \ ^\circ C^{-1} \times 2.7 ^\circ C = 1130.22 \ J$

I.

$\Delta T$ for room temp water

For Trial 1

$\Delta T_1 = |T(after \ mixing) - T(hot \ water)| = |19.5 - 20.2| =0.7 ^\circ C$

For Trial 2

$\Delta T_1 = |T(after \ mixing) - T(hot \ water)| = |21.8 - 22.2| =0.4 ^\circ C$

J.

Mass of room temp water.

For trial 1

Volume of room temp water = 75 mL

Density = 1g/mL

hence, mass = Volume* Density = 75 mL * 1 g/ mL = 75 g

For Trial 2

Volume of room temp water = 100 mL

Density = 1g/mL

hence, mass = Volume* Density = 100 mL * 1 g/ mL = 100 g

K.

For Trial 1

Heat gained by room temp water is given by

$q_{hot}= m \times C \times \Delta T = 75 \ g \times 4.186 \ J \ g^{-1} \ ^\circ C^{-1} \times 0.7 ^\circ C = 209.765 \ J$

For Trial 2

$q_{hot}= m \times C \times \Delta T = 100 \ g \times 4.186 \ J \ g^{-1} \ ^\circ C^{-1} \times 0.4 ^\circ C = 167.44\ J$

L.

The heat lost by the hot water is gained by both the room temp water and calorimeter.

hence,

$q_{cal} + q_{rt} = q_{hot} \\ \Rightarrow q_{cal} = q_{hot} - q_{rt}$

For Trial 1

For Trial 2

M.

The temperature change for the calorimeter is same as that of the room temp water.

Also,

$q_{cal}= C_{cal} \times \Delta T \\ \Rightarrow C_{cal} = \frac{q_{cal}}{\Delta T}$

For Trial 1

$\Delta T = 3.8 ^\circ C$

$C_{cal} =\frac{983.245 \ J}{3.8 \ ^\circ C} = 258.74 \ J/ ^\circ C$

For Trial 2

$\Delta T = 2.7 ^\circ C$

$C_{cal} =\frac{962.78 \ J}{2.7 \ ^\circ C} =356.58 \ J/ ^\circ C$

N

Average C_cal =

$\frac{258.74+356.58}{2} = 307.66 \ J/ ^\circ C$