Worksheet- Calculations involving Specific Heat

1. For q= m ●c ● Δ T : identify each variables by name & the units associated with it.

q = amount of heat (J)

m = mass (grams)

c = specific heat (J/g°C)

ΔT = change in temperature (°C)

2. Heat is not the same as temperature, yet they are related. Explain how they differ from each other.

Heat is a combination of kinetic energy (measured by temperature) and potential energy.

a. Perform calculations using: (

q= m

●

Δ T) b. Determine if it’s endothermic or exothermic

1. Gold has a specific heat of 0.129 J/(g×°C). How

many joules of heat energy are required to raise

the temperature of 15 grams of gold from 22 °C to 85

°C?

q = ?

m = 15 g

c = 0.129 J/g°C

ΔT = 85°C - 22°C = 63°C

q = mc ΔT

q = (15 g)(0.129 J/g°C)(63 °C)

q = 121.9 J  120 J

Endothermic or exothermic? Endothermic

2. An unknown substance with a mass of 100 grams

absorbs 1000 J while undergoing a temperature increase

of 15 °C. What is the specific heat of the substance?

q = 1000 J

m = 100 g

c = ?

ΔT = 15°C

c = 0.666667 J/g°C  0.7 J/g°C

Endothermic or exothermic? Endothermic

3. If the temperature of 34.4 g of ethanol increases

from 25 °C to 78.8 °C, how much heat has been absorbed

by the ethanol? The specific heat of ethanol is 2.44

J/(g×°C)

q = ?

m = 34.4 g

c = 2.44 J/g°C

ΔT = 78.8°C - 25°C = 53.8°C  54 °C

q = (34.4 g)(2.44 J/g°C)(54 °C)

q = 4532.544 J  4500 J

Endothermic or exothermic? Endothermic

4. Graphite has a specific heat of 0.709 J/(g×°C). If a

25 gram piece of graphite is cooled from 35 °C to 18 °C,

how much energy was lost by the graphite?

q = ?

m = 25 g

c = 0.709 J/g°C

ΔT = 18°C - 35°C = -17°C

q = (25 g)(0.709 J/g°C)(-17 °C)

q = --301.325 J  -300 J

Endothermic or exothermic? Exothermic

5. Copper has a specific heat of 0.385 J/(g×°C). A piece

of copper absorbs 5000 J of energy and

undergoes a temperature change from 100 °C to

200 °C. What is the mass of the piece of copper?

q = 5000 J

m = ?

c = 0.385 J/g°C

ΔT = 200°C - 100°C = 100°C

m= 129.87 g  100 g

Endothermic or exothermic? Endothermic

6. 45 grams of an unknown substance undergoes a

temperature increase of 38 °C after absorbing

4172.4 Joules. What is the specific heat of the

substance? Look at the table on page 513 of your

book, and identify the substance.

q = 4172.4 J

m = 45 g

c = ?

ΔT = 38°C

c = 2.44 J/g°C  2.4 J/g°C  Ethanol

Endothermic or exothermic? Endothermic

Partial preview of the text

Download Worksheet Calculations involving Specific Heat and more Exercises Chemical Thermodynamics in PDF only on Docsity!

1. For q= m ●c ● Δ T : identify each variables by Worksheet-^ Calculations involving Specific Heat c = specific heat (J/g q = amount of heat (J) m = mass (grams) name & the units°C associated with it).

a. (^) Heat is not the same as temperature, yet they are related. Heat is a combination of kinetic energy (measured by temperature) and potential energy. Perform calculations using: ( q= m ●^ ΔT = change in temperature ( c ● Δ T) b. Determine if it’s endothermic or exothermic Explain how the°C) y differ from each other.
1. many joules of heat energy are required to raise the temperature of 15 grams of gold from 22 ° q = ?C? Gold has a specific heat of 0.129 J/(g×°C). How °C to 85 m = 15 g c = 0.129 J/g ΔT = 85 q = mc ΔT°C - 22 °C°C = 63°C q = (15 g)(0.129 J/g q = 121.9 J Endothermic  or exothermic? 120 J °C)(63 °C ) Endothermic
  1. absorbs 1000 J while undergoing a temperature increase of 15 q = 1000 J m = 100 g An unknown substance with a mass of 100 grams °C. What is the specific heat of the substance? c =? ΔT = 15°C
2. from If the temperature of 25 °C to 78.8 °C, how much heat has been absorbed 34.4 g of ethanol increases^ c = 0.666667 J/g^ Endothermic or exothermic?°C^ ^ 0.7 J/g°C^ Endothermic by the ethanol? J/(g q = m = c = ?2.4434.4×°C) J/gg °C The specific heat of ethanol is 2. ΔT = q = ( q = 4532.54434.4 78.8 g)(°C2.44 - J (^2 5)  °CJ/g 4500 = (^) °C53.8)( (^54) J °C °C ) 54 °C Endothermic or exothermic? Endothermic
  1. 25 gram piece of graphite is co how much energy was lost by the graphite? q = m = Graphite has a specific heat of 0.709 J/(g? 2 5 g oled from 35 °×°C to 18C). If a °C, c = ΔT = q = ( q = 0.709-- (^2) 301.325 (^18) 5 g)(0.°C J/g - 70935 °CJ°C J/g = - - (^17) °C 300 J °C)(- 17 °C)
3. of copper absorbs 5000 J of energy and undergoes a temperature change from 100 200 Copper has a specific heat of °C. What is the mass of the piece of copper? 0.385 J/(g °×°C toC). A piece^ Endothermic or exothermic?^ Exothermic q = m = c = 0. ΔT = 5000 J? 385200 °CJ/g - °C 100 °C = 100 °C m Endothermic or exothermic?= 129.87 g  100 g Endothermic
  1. temperature increase of 38 4172.4 Joules. What is the specific heat of the substance? book, and identify the substance. 45 grams of an unknown substance undergoes a Look at the table on pa °C after absorbing ge 513 of your q = m = c = ΔT = 4172.4 J? 4 5 g 38 °C c = Endothermic or exothermic? 2.44 J/g°C  2.4 J/g°C (^) Endothermic  Ethanol

How much did the water temperature change specific heat of water is 4.18 J/(g q = m = A 500 J 40 40 g g sample of water absorbs 500 Joul×°C). es of energy.? The c = 4.18 ΔT =? J/g°C Endothermic or exothermic?^ =^ 2.99 °C^ ^ 3°C Endothermic
1. what is the final temperature of the water? Liquid water has a specific heat of 4.18 J/(g q = m = If 335 g of water at 65.5 - 335 9750 J g °C loses 9750 J of heat,×°C). c = ΔT = 4.18 Tfinal J/g – °C65.5 °C

Worksheet Calculations involving Specific Heat, Exercises of Chemical Thermodynamics

Partial preview of the text

Download Worksheet Calculations involving Specific Heat and more Exercises Chemical Thermodynamics in PDF only on Docsity!

1. For q= m ●c ● Δ T : identify each variables by Worksheet-^ Calculations involving Specific Heat c = specific heat (J/g q = amount of heat (J) m = mass (grams) name & the units°C associated with it).

2. Endothermic or exothermic?^ Exothermic