10 Questions
| Attempts: 541

Heat absorbed = (mass) (Specific Heat Capacity Cp) (change in temperature)

Questions and Answers

- 1.If 80 grams of water is heated from 26
^{o}C to 31^{o}C, how much heat is absorbed? (C_{p}of liquid water = 4.184 J/g^{o}C) (Heat absorbed/lost q) m = (mass m ) (change in temperature ∆ t) (specific heat capacity C_{p})- A.
400 J

- B.
1,673.6 J

- C.
1,500 J

- D.
1,338.9

- 2.A piece of aluminum (3.6 g) is heated from 20
^{o}C to 30^{o}C. If the specific heat of aluminum os 0.897 J/g^{o}C, how much heat energy was absorbed?- A.
322.92 J

- B.
32.292 J

- C.
3.2292

- D.
3.6 J

- 3.A 34 gram piece of an unknown metal absorbs 351.56 Joules of energy when the temperature increased from 10
^{o}C to 32^{o}C.*What is the specific heat of the substance?*Hint: You are solving for Specific Heat (C_{p})*not*heat absorbed.- A.
.47 J/gC

- B.
.3 J/gC

- C.
45.2 J/gC

- D.
3.229 J/gC

- 4.A 33.3 gram piece of iron (Cp 0.449 J/gC) absorbs 290.0 J of energy. What was the increase (change in temperature) in degrees celsius?
- A.
3.9 C

- B.
10.4 C

- C.
19.4 C

- D.
15.4 C

- 5.A 6.8 g sample of glass was heated from 4 C to 39 C. It was found to have absorbed 59.5 J of energy. What is the specific heat of this type of glass?
- A.
35 J/g C

- B.
306 J/g C

- C.
.35 J/g C

- D.
.25 J/g C

- 6.A 42.6 gram pie ce of metal is heated to a temperature of 100 C. A 250 ml container of water (recall 1 g of water = 1 ml) starts at a temperature of 17 C. When the metal is added to the water, the water increases in temperature to 20 C. (The specific heat capacity of water is 4.184 J/g C) What is the specific heat capacity of the metal? Assume a closed system. How to aproach the problem: Write what you know about the water and about the metal: Water: Change in temperature 17 to 20; 250 g; 4.184 J/gC Metal: Change in temperature 100 to 20; 42.6g ; Cp unknown Recall that the heat lost by the metal will equal the heat gained by the water in a closed system. Since the change in temperature, mass and specific heat capacity of water are all known, solve for heat abosrbed by water first: Heat absorbed by water= (3 C) (250) (4.184). This answer will be the SAME as the heat lost by the metal. 3,138 J of heat lost by water (or gained by metal). Now go to metal side. Use the 3,138 J from the water equation for the heat lost from the metal. 3,138 J = (100-20 C) (42.6 g) (Cp) .
*Now solve for the Cp of metal*- A.
.44 J/g C

- B.
.92 J/g C

- C.
.56 J/g C

- D.
.85 J/gc

- 7.A 96.85 g piece of metal is heated to a temperature of 100 C. A 250 ml container of water( recall: 1 g of water = 1 ml of water) metal is added to the water, the water increases in temperature from 13 C to 19 C. (The specific heat capacity of water is 4.184 J/g C). What is the specific heat capacity of the metal? Assume a closed system. How to approach this problem: Write what you know about water and about the metal. Remember that the heat lost from the metal is gained by the water. Solve frist for the heat absorbed/loss on the side without unknowns (i.e. water).
- A.
.8 J/g C

- B.
.56 J/g C

- C.
1.2 J/g C

- D.
.66 J/g C

- 8.Whichtemperature represents the heat of vaporization? (Hint: Change in state: (Ex: Solid to liquid; liquid to gas) will NOT have an increase in temperature.
- A.
A

- B.
B

- C.
F

- D.
G

- 9.Which area on this graph represents a substance that is completely a gas?
- A.
B

- B.
C

- C.
D

- D.
E

- 10.If 100 grams of water is heated from 19 C to 23 C how much heat is absorbed? Cp water = 4.184 J/gC
- A.
.2 J

- B.
120 J

- C.
1,673.6 J

- D.
1,199 J

×

Wait!

Here's an interesting quiz for you.