Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Can you help me number 4? Is this two correct answer with show my work including the Significant Figures number?
Transcribed Image Text:Calculations for Temperature and Phase Change Worksheet
The heat of fusion of ice is 79.7 cal/g.
The heat of vaporization of water is 540 cal/g.
Report the answer using the correct number of significant figures!
1. How much energy is required to melt 100.0 grams of ice?
100.0 g 79.7 cal
·cal = 7970 cal
9
2. How much energy is required to vaporize 234.5 g of water?
1.2663 х 105
234.5g 540 cal
g
1.3X105cal
Answer: 1.3 x 105 cal
3. If 30.6 calories are required to vaporize 25 g of a substance, what is the heat of vaporization of
that substance?
30.6 cal
259
1.224
1.2 cal/g
Answer: 1.2 cal/g
4. How much energy is removed from 500.0 g of water when the temperature is lowered by 1.10 °C?
500.09
1 cal) (-1.10°C) = [550 cal
500.g
-2,3012×103
4.184J (-1.10) = √2.30×10³5
Answer: -550. cal (or -2.30 x 10³ J)
I cal
5. How much energy is required to raise the temperature of 1000.0 g of water from 23.00 °C to
M=1000.09 Q=S•m. (OT)
s=lcal/g °C
T1=23.00%
26.00 °C?
1000,03 (1Cal) (3°C) = 5+00X10³ cal
go c
1,2552x10
-1.26x1045
3,00x10³ cal 4.1845).
1461845-1-20
10.0924 cal
9°C
Answer: 7970 cal
1) (100.0°C) = 12,4 cal
6. The specific heat of copper is (0.0924 cal/g°C), how much energy is required to raise the
temperature of 10.0 g of copper by 100.0 °C?
10.09
T2= 26.00°C
Answer: 3.00 x 10³ cal (or 1.26 x 10¹J)
25,65=786g xs x (35°C -20°c)
5=25.65
2.171331637X10
(7869 x 15°C) (2.17X/10-3³5/9°C
Tfinal-Tinitial
26.00-23.00=3c
-3
Answer: 92.4 cal
7. If 25.6 J of energy raised 786 g of a substance from 20.0°C to 35.0°C, what is the specific heat of
the substance (S)?
Tfinal-Tinitial
35.0°C-20.0°= 15.00
Answer: 2.17 x 10-³J/g °C
Expert Solution
Step 1
Amount of heat,Q required to change the temperature of a substance of mass ,m and specific heat capacity, C from temperature T1 to temperature T2 is calculated by formula :
Q = m C ( T2 - T1 )
Q = m C ( ∆T )
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