**Prequiz
14.1 - Heat and Calorimetry .:. Go
Up**

The Formulas:

Q = heat added in Joules

m = mass in kg

c = specific heat in J/kg^{o}C. THis is the heat needed to raise one kg one^{o}C

Q = heat added in Joules

m = mass in kg

L = Latent heat of fusion (melting) or vaporization (boiling) in J per kg

The Prequiz:

**Show your work, and circle
your answers and use sig figs to receive full credit.**

This is a graph of temperature v. heat added for a .218 kg sample of unknown. It starts as a solid, and ends as a gas.

** **

1. What is the **melting
temperature** and the **boiling temperature** of this substance? Label
the **solid**, **liquid** and **gaseous** phases.

(Melt = 30 ^{o}C,
Boil = 50 ^{o}C, solid 0-300 J added, Liquid 700-1100 J added, Gas
1700-2000 J added.)

Where the temperature is rising, it is all one phase (i.e. solid, liquid, gas). Where the temperature plateaus, the substance is undergoing a phase change - either melting, or boiling.

**2. What is the latent heat
of fusion? (Melting)**

(1830 J/kg)

Use

Q = 400 J (starts at 300 J, ends at 700 J)

m = 0.218 kg (see above the graph)

L = 1834.86 =1830 J/kg

**3. What is the specific
heat of the liquid phase?**

(91.7 J/kg^{o}C)

Use

Q = 400 J (starts at 700 J, ends at 1100 J - Read the graph)

m = 0.218 kg (see above the graph)

DT = 20^{o}C (starts at 30^{o}C ends at 50^{o}C - Just read the graph)

C = 91.743 =91.7 J/kg^{o}C

4. What heat do you need to
heat 3.29 Kg of water at 21.0 ^{o}C to steam at 175 ^{o}C?
(For H_{2}O: C_{ice} = 2100 J/kg^{o}C, l_{f} =
3.33 x 10^{5} J/Kg, C_{water} = 4186 J/Kg^{o}C, l_{v}
= 22.6 x 10^{5 }J/Kg, C_{steam} = 2010 J/kg^{o}C)

(9,019,350.76 J – 9.02 x 10^{6}
J)

This will need to heat to
boiling from 21.0 ^{o}C to 100.0 ^{o}C (The boiling point of
water), then turn to steam, then heat from 100.0 ^{o}C to 175 ^{o}C.
Add all the heats up.

First - use

m = 3.29 kg

DT = 79^{o}C (starts at 21.0^{o}C ends at 100^{o}C )

C = 4186 J/kg^{o}C (for the liquid)Q = 1,087,983.26 J

Next use

m = 3.29 kg

L = 22.6 x 10^{5 }J/Kg (vaporization - it's boiling)Q = 7,435,400 J

Next - use

m = 3.29 kg

DT = 75^{o}C (starts at 100^{o}C ends at 175^{o}C )

C = 2010 J/kg^{o}C (steam)Q = 495,967.5 J

Finally - add them all up:

1,087,983.26 J + 7,435,400 J + 495,967.5 J = 9,019,350.76 J =9.02 x 10^{6}J

5. 500. grams of a mystery
liquid at 45 ^{o}C is mixed with 300. grams of water (C = 4186 J/Kg^{o}C)
initially at 22 ^{o}C. The final temperature of the mixture is 33 ^{o}C.
What is the specific heat of the mystery liquid? (Assuming no heat was lost to
the surroundings)

(2,302.3 – 2.30 x 10^{3}
J/kg^{o}C)

Calorimetry problems are

m_{1}c_{1}DT_{1}= m_{2}c_{2}DT_{2 }

Sometimes there is more than one term on a side, but the left side is for the hot things that lose heat, and the right side is the things that gain the heat:

m= .500 kg_{1 }

c= ??_{1 }

DT= 45 - 33 = 12_{1 }^{o}C

m= 0.300 kg_{2 }

c= 4186 J/kg_{2 }^{o}C

DT= 33 - 22 = 11_{2 }^{o}C