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4. How much energy is removed from 500.0 g of water when the temperature is lowered by 1.10 °C? cal

4. How much energy is removed from 500.0 g of water when the temperature is lowered by 1.10 °C? cal

Chemistry
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?

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
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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