# specific heat capacity questions and answers 1

Define specific heat capacity and derive the mathematical formula.

# Answer

The specific heat capacity is defined as the quantity of heat required to raise the temperature of a unit mass (1K) of a body through a degree rise in temperature(1k).

The formula of the specific heat capacity. Note that the quantity of heat a body receives is directly proportional to it mass, change in temperature, and also the nature of the material. That is:

Q ∝ m(Ѳ_{2}-Ѳ_{1}) or Q = mc(Ѳ_{2}-Ѳ_{1}) therefore

Where Q = the quantity of heat, m = unit mass of the body, c = specific heat capacity and (Ѳ_{2}=Ѳ_{1}) =change in temperature.

Questions and answers on elasticity

# specific heat capacity questions and answers 2

what is heat capacity?

# Answer

Heat capacity, can defined as the quantity of heat required to raise the temperature of the entire body through a degree rise in temperature. (1K). and is given as

**h = cm **

Where h = heat capacity, c = specific heat capacity and m = the mass of the body.

# specific heat capacity questions and answers 3

Give that the specific heat capacity of iron as 450JKg^{-1}K^{-1}, and the heat capacity is 1800JK^{-1}, calculate the mass of the iron.

# Answer

**h =cm**

where H= heat capacity =1800JK^{-1}, c= specific heat capacity = 450Kg^{-1}K^{-1}, so

900 = 450m

m = 1800/450 = 4kg

the mass of the iron = 4.0kg.

Questions and answers on momentum

# specific heat capacity questions and answers 4

what quantity of heat energy is needed to raise the temperature of 20kg 0f aluminum through a through 10k (specific heat capacity of aluminum = 900JKg^{-1}K^{-1})

# Answer

**Q = mc(Ѳ _{2}-Ѳ_{1}) **

where Q = the quantity of heat, m= mass = 20kg and (Ѳ_{2}-Ѳ_{1}) =change in temperature = 10K, c = 900JKg^{-1}K^{-1} ) so

Q = 20 x 900 x 10 = 180000

Q =18 x 10^{4}J

# specific heat capacity questions and answers 5

Calculate the heat capacity of 20kg of brass ( specific heat capacity of brass = 380Jkg^{-1}K^{-1})

# Answer

**h =cm**

, where h = heat capacity , m= 20kg, c 380JKg^{-1}K^{-1} therefore

h = 20 x 380 = 7600JK^{-1}

Heat capacity = 7600JK^{-1}

# specific heat capacity questions and answers 6

A piece of iron of specific heat capacity 0.04JKg^{-1}K^{-1} and mass 400kg, is quickly dropped into 30kg of water at 10^{o}c contain in a calorimeter of 120kg and specific heat capacity of 0.1JKg^{-1}K^{-1}. If the temperature of the mixture is 30^{o}c, calculate the initial temperature of the hot iron (4200JKg^{-1}K^{-1} = specific heat capacity of water)

Questions and answers on simple machine

# Answer

To solve this you make heat loss by the iron = heat gain by the calorimeter and water.

**m _{i}c_{i} (Ѳ_{3}-Ѳ_{2}) = m_{w}c_{w} (Ѳ_{2}-Ѳ_{1}) + m_{c}c_{c} (Ѳ_{2}-Ѳ_{1})**

400 x 0.04 x (Ѳ_{3}-Ѳ_{2} ) = 30 x 4200 (Ѳ_{2}-Ѳ_{1} ) + 120 x 0.1 x(Ѳ_{2}-Ѳ_{1} )

16(Ѳ_{3}-30 ) = 126000 x (30-10) + 12 (30-10 )

16Ѳ_{3} – 480 = 2520000 + 240

16Ѳ_{3} =2520000 + 240 + 480

16Ѳ_{3} = 2520720

The initial temperature of the iron 157545K.

# specific heat capacity questions and answers 7

A liquid of specific heat capacity of 3000JKg^{-1}K^{-1} rise from 15oc to 65^{o}c in one minute when an electric heater is used. If the heater generates 63000J per minutes, calculate the mass of the water.

# Answer

**cm (Ѳ _{2-} Ѳ_{1}) = ivt **

where c = specific heat capacity of the liquid= 3000JKg^{-1}K^{-1},

m = mass of the liquid,

(Ѳ_{2-} Ѳ_{1}) = change in the temperature of the liquid=( 65 -15) = 50^{o}c

and ivt = the quantity of energy generated by the heater = 63000J

**cm (Ѳ _{2-} Ѳ_{1}) = ivt =**

3000 x 50 x m = 63000

150000m = 63000

m = 63000/150000 = 0.42

m =0.42kg

Questions and answers on uncertainty principle

# specific heat capacity questions and answers 8

A certain metal of mass 1.5kg at initial temperature of 27^{o}c, absorbed heat from electric heater of 75watts of power rating for 4 minutes . If the final temperature was 47^{o}c , calculate the specific heat capacity of the metal and it heat capacity.

# Answer

**Ivt = cm (Ѳ _{2-} Ѳ_{1})**

Iv = 75watts, ivt = 75 x 4 x 60 =18000J, m = 1.5kg, (Ѳ_{2-} Ѳ_{1})= 47^{o}c -27^{o}c = 20^{o}c and c =?

**Ivt = cm (Ѳ _{2-} Ѳ_{1})**

1800 = 1.5 x 20 x c

18000 = 30c

C = 18000/30 = 600Jkg^{-1}k^{-1}

To find the heat capacity

H = cm where H = heat capacity, c = specific heat capacity = 600Jkg^{-1}k^{-1} and m -1.5kg

H = 600 x 1.5 = 900JK^{-1}

# specific heat capacity questions and answers 9

What is specific latent heat of fusion?

# Answer

Specific latent heat of fusion is the quantity of heat required to convert a unit mass of solid to liquid without change in its temperature.

# specific heat capacity questions and answers 10

what is specific latent heat of evaporation?

# Answer

Specific latent heat evaporation can be defined as the quantity of heat needed to convert a unit mass of liquid to gas without changing it temperature.

# specific heat capacity questions and answers 11

An electric current of 3A flowing through an electric heating element of resistance 2Ω embedded in 1000g of an oil, raise the temperature of the oil by 10^{o}c in 10 seconds. Calculate the specific heat capacity of the oil.

# Answer

V = IR

V= potential difference in volts. I = current = 3A and R = 2Ω

V = 3 x 2 = 6volts

**Ivt = cm (Ѳ _{2-} Ѳ_{1})**

Ivt = 3 x 6 x 10 = 180J, m = 1000g = 1Kg and (Ѳ_{2-} Ѳ_{1}) = 10^{o}c

**Ivt = cm(Ѳ _{2-} Ѳ_{1})**

180 = 1 x 10 x c

180 = 10c

C = 180/10 = 18

C = 18JKg^{-1}c^{-1} (specific heat capacity of the oil)

# specific heat capacity questions and answers 12

.How much heat energy is required to change 10kg of ice to water at 0^{o}c? Take specific latent heat of fusion = 336000JKg^{-1}K^{-1}.

# Answer

Q = ml, where Q = quantity of heat required =? m = mass of the ice = 10kg and specific latent heat of fusion of ice = 336000JKg^{-1}K^{-1}

**Q = ml**

Q = 10 x 336000 = 3360000

Q = 3360000J

# specific heat capacity questions and answers 13

If 1.13 x 10^{6}J of heat energy is required to convert 15kg of steam to water , calculate the specific latent heat of vaporization

# Answer

Q = 1.13 x 10^{6}J, m=15kg, and L =?

**Q = ml**

1.13 x 10^{6} = 15 x L

L =75333.3JKgk^{-1}K^{-1}

# specific heat capacity questions and answers 14

what quantity of heat is needed to melt 120g of ice at -20^{o}c to water at 30^{o}c? Specific heat capacity of water and ice are 4200JKg^{-1}K^{-1} and 2100JKg^{-1}K^{-1} respectively. And also to specific latent heat 0f ice as 3.36 x 10^{5}JK^{-1}

# Answer

To solve this, it is going to take three stages before the final answer.

(a). the quantity of heat needed to change the ice of 120g from -20^{o}c to 0^{o}c

**Q = c _{i}m (Ѳ_{2-} Ѳ_{1})**

C_{i} =2100JKg^{-1}K^{-1}, m = 120g = 0.12kg, and (Ѳ_{2-} Ѳ_{1}) = 0-(-20) = 20^{o}c

Q = 2100 x 0.12 x 20 = 5040J

(b). the quantity of heat need to melt ice of 0^{o}c to water of 0^{o}c

**Q = mL**

m = 0.12kg, and L = 3.36 x 10^{5}JK

Q = 0.12 x 3.36 x 10^{5} = 40320J

(c). the quantity of heat needed to change the water from 0oc to 30oc

**Q = mc(Ѳ _{2-} Ѳ_{1})**

m =0.12kg, c = 4200JKg^{-1}K^{-1} (specific heat capacity of water), and (Ѳ_{2-} Ѳ_{1})= 30 -0 = 30^{o}c

Q = 0.12 x 4200 x 30 =15120J

So quantity of heat is needed to melt 120g of ice at -20^{o}c to water at 30^{o}c, is sum total of a, b, and c

Q= 5040 + 40320 + 15120 = 60480J

# specific heat capacity questions and answers 15

A 30v electric heater is used to supply a current of 10A for 1200 seconds to a mass of 1kg, the body melts and rises through a temperature of 50^{o}c, at an extra one minute. Determine latent heat of fusion/ and specific heat capacity of the solid.

# Answer

**Ivt = mL**

m = 1kg, I = 10A, v = 30v, t = 1200s and L = latent heat of fusion

10 x 30 x 1200 = 1 x L

360000 = L

L =360000JK^{-1} (latent heat of fusion)

To calculate the specific heat capacity of solid, then you calculate the total heat needed to melt and raise the temperature 0oc to 50oc at an extra one minute.

Ivt = mL +mc(Ѳ_{2-} Ѳ_{1})

I = 10A, m =ikg, L =360000JK-1, v = 30v, t = 1200 + 0ne minute = 1200 + 60 = 1260s (Ѳ_{2-} Ѳ_{1}) = 50oc and c = specific heat capacity of the solid

10 x 30 x 1260 = 36000 + 1 x 50 x c

378000 = 360000 + 50c

50c = 378000-360000 =1800

C = 18000/50 = 360JKg^{-1}K^{-1}

Specific heat capacity == 3.6 x 10^{2}JKg^{-1}K^{-1}

Specific latent heat = 3.6 x10^{5}JK^{-1}

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