Postlab for Calorimetry SD Flashcards


Set Details Share
created 7 months ago by BerryJamJam
27 views
Attempt 1 and Attempt 2
updated 7 months ago by BerryJamJam
Subjects:
general chemistry i lab
show moreless
Page to share:
Embed this setcancel
COPY
code changes based on your size selection
Size:
X
Show:

1

A metal sample weighing 121.00 grams and at a temperature of 98.6 degrees Celsius was placed in 42.00 grams of water in a calorimeter at 24.0 degrees Celsius. At equilibrium the temperature of the water and metal was 35.2 degrees Celsius.

What is the change in temperature of the water (in degrees C)? (Report this as a positive value)

T = Tf - Ti

T = 35.2 - 24.0 = 11.2 degrees C

2

A metal sample weighing 126.00 grams and at a temperature of 96.9 degrees Celsius was placed in 48.00 grams of water in a calorimeter at 22.0 degrees Celsius. At equilibrium the temperature of the water and metal was 38.4 degrees Celsius.

What is the change in temperature of the metal (in degrees C)? (Report this as a positive value)

T = Tf - Ti

T = 38.4 - 96.9 = -58.5 degrees C -> 58.5 degrees C

3

A metal sample weighing 130.00 grams and at a temperature of 96.5 degrees Celsius was placed in 47.50 grams of water in a calorimeter at 24.5 degrees Celsius. At equilibrium the temperature of the water and metal was 43.6 degrees Celsius.

How much heat flowed into the water? (Report your answer in J)

The specific heat of the water is 4.184 J/g/C

Report your answer to three significant figures. The value should be positive.

DO NOT USE EXPONENTIAL NOTATION

q = m Cs △T

△T = Tf - Ti

△T = 43.6°C - 24.5°C = 19.1°C

q = (47.50 g) x (4.184 J/g°C) x (19.1°C)

q = 3,795.9 J

4

A metal sample weighing 123.50 grams and at a temperature of 96.1 degrees Celsius was placed in 47.50 grams of water in a calorimeter at 22.8 degrees Celsius. At equilibrium the temperature of the water and metal was 40.3 degrees Celsius.

Calculate the specific heat of the metal (in J/g/C).

The specific heat of the water is 4.184 J/g/C

Report your answer to three significant figures.

DO NOT USE EXPONENTIAL NOTATION

47.50 g (4.184 J/g°C) (40.3°C - 22.8°C) = (123.50 g) (c) (96.1°C - 40.3°C)

198.74 (17.4) = (123.50 g) (c) (55.8)

3477.95 = 6891.3 c

0.505 J/g/C = c

5

A metal sample weighing 128.00 grams and at a temperature of 98.2 degrees Celsius was placed in 44.00 grams of water in a calorimeter at 22.5 degrees Celsius. At equilibrium the temperature of the water and metal was 36.1 degrees Celsius.

What is the change in temperature of the water (in degrees C)? (Report this as a positive value)

T = Tf - Ti

T = 36.1 - 22.8 = 13.3 degrees C

6

A metal sample weighing 127.50 grams and at a temperature of 98.2 degrees Celsius was placed in 42.00 grams of water in a calorimeter at 22.0 degrees Celsius. At equilibrium the temperature of the water and metal was 42.5 degrees Celsius.

What is the change in temperature of the metal (in degrees C)? (Report this as a positive value)

T = Tf - Ti

T = 42.5 - 98.2 = -55.7 degrees C -> 55.7 degrees C

7

A metal sample weighing 126.00 grams and at a temperature of 98.2 degrees Celsius was placed in 44.50 grams of water in a calorimeter at 22.7 degrees Celsius. At equilibrium the temperature of the water and metal was 35.8 degrees Celsius.

How much heat flowed into the water? (Report your answer in J)

The specific heat of the water is 4.184 J/g/C

Report your answer to three significant figures. The value should be positive.

DO NOT USE EXPONENTIAL NOTATION

q = m Cs △T

△T = Tf - Ti

△T = 35.8°C - 22.7°C = 13.1°C

q = (44.50 g) x (4.184 J/g°C) x (13.1°C)

q = 2439.1 J

8

A metal sample weighing 124.50 grams and at a temperature of 97.2 degrees Celsius was placed in 47.00 grams of water in a calorimeter at 24.7 degrees Celsius. At equilibrium the temperature of the water and metal was 39.8 degrees Celsius.

Calculate the specific heat of the metal (in J/g/C).

The specific heat of the water is 4.184 J/g/C

Report your answer to three significant figures.

DO NOT USE EXPONENTIAL NOTATION

47.00 g (4.184 J/g°C) (39.8°C - 24.7°C) = (124.50 g) (c) (97.2°C - 39.8°C)

196.648 (15.1) = (124.50 g) (c) (57.4)

2969.3848 = 7088.9 c

0.416 J/g/C = c