# specific heat capacity questions and answers 1

Define specific heat capacity and derive the mathematical formula.

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(Ѳ21)   or  Q  = mc(Ѳ21)   therefore $c\quad =\quad \frac { Q }{ m({ { \theta }_{ 2 }{ -\theta }) } }$

Where Q = the quantity of heat, m = unit mass of the body, c = specific heat capacity and (Ѳ21) =change in temperature.

# specific heat capacity questions and answers 2

what is heat capacity?

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-1K-1, and the heat capacity is 1800JK-1, calculate the mass of the iron.

h =cm

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

900 = 450m

m = 1800/450 = 4kg

the mass of the iron = 4.0kg.

# 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-1K-1)

Q  = mc(Ѳ21)

where  Q = the quantity of heat, m= mass = 20kg and (Ѳ21) =change in temperature = 10K, c = 900JKg-1K-1 ) so

Q = 20 x 900  x 10 = 180000

Q =18 x 104J

# specific heat capacity questions and answers 5

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

h =cm

, where h = heat capacity , m= 20kg, c 380JKg-1K-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-1K-1 and mass 400kg, is quickly dropped into 30kg of water at 10oc contain in a calorimeter of 120kg and specific heat capacity of 0.1JKg-1K-1. If the temperature of the mixture is 30oc, calculate the initial temperature of the hot iron (4200JKg-1K-1 = specific heat capacity of water)

Questions and answers on simple machine

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

mici32) = mwcw21) + mccc21)

400 x 0.04 x (Ѳ32 )  = 30 x 4200 (Ѳ21 ) + 120 x 0.1 x(Ѳ21 )

16(Ѳ3-30 ) = 126000  x (30-10) + 12 (30-10 )

16Ѳ3 – 480 = 2520000 + 240

16Ѳ3 =2520000 + 240 + 480

16Ѳ3 = 2520720 ${ \theta }_{ 3 }=\frac { 2520720 }{ 16 } \quad =157545K$

The initial temperature of the iron 157545K.

# specific heat capacity questions and answers 7

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

cm (Ѳ2- Ѳ1) = ivt

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

m = mass of the liquid,

2- Ѳ1) = change in the temperature of the liquid=( 65 -15) = 50oc

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

cm (Ѳ2- Ѳ1) = 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 27oc, absorbed heat from electric heater of 75watts of power rating for 4 minutes . If the final temperature was 47oc , calculate the specific heat capacity of the metal and it heat capacity.

Ivt = cm (Ѳ2- Ѳ1)

Iv  = 75watts,  ivt = 75 x 4 x 60 =18000J, m = 1.5kg,  (Ѳ2- Ѳ1)= 47oc -27oc = 20oc and c =?

Ivt = cm (Ѳ2- Ѳ1)

1800 = 1.5 x 20 x c

18000  =  30c

C = 18000/30 = 600Jkg-1k-1

To find the heat capacity

H = cm where H = heat capacity, c = specific heat capacity = 600Jkg-1k-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?

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?

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 10oc in 10 seconds. Calculate the specific heat capacity of the oil.

V = IR

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

V = 3 x 2 = 6volts

Ivt = cm (Ѳ2- Ѳ1)

Ivt = 3 x 6 x 10 = 180J, m = 1000g = 1Kg and (Ѳ2- Ѳ1) = 10oc

Ivt = cm(Ѳ2- Ѳ1)

180 = 1 x 10 x c

180 = 10c

C = 180/10 = 18

C = 18JKg-1c-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 0oc? Take specific latent heat of fusion = 336000JKg-1K-1.

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

Q  = ml

Q = 10 x 336000 = 3360000

Q  = 3360000J

# specific heat capacity questions and answers 13

If 1.13 x 106J of heat energy is required to convert 15kg of steam to water , calculate the specific latent heat of vaporization

Q = 1.13 x 106J, m=15kg, and L =?

Q = ml

1.13 x 106 = 15 x L

L =75333.3JKgk-1K-1

# specific heat capacity questions and answers 14

what quantity of heat is needed to melt 120g of ice at -20oc to water at 30oc? Specific heat capacity of water and ice are 4200JKg-1K-1 and 2100JKg-1K-1 respectively.  And also to specific latent heat 0f ice as 3.36 x 105JK-1

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 -20oc to 0oc

Q = cim (Ѳ2- Ѳ1)

Ci =2100JKg-1K-1,  m = 120g = 0.12kg, and (Ѳ2- Ѳ1) = 0-(-20) = 20oc

Q = 2100 x 0.12 x 20 = 5040J

(b). the quantity of heat need to melt ice of 0oc to water of 0oc

Q = mL

m = 0.12kg, and L = 3.36 x 105JK

Q = 0.12 x 3.36 x 105 = 40320J

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

Q = mc(Ѳ2- Ѳ1)

m =0.12kg, c = 4200JKg-1K-1 (specific heat capacity of water), and (Ѳ2- Ѳ1)= 30 -0 = 30oc

Q = 0.12 x 4200 x 30 =15120J

So quantity of heat is needed to melt 120g of ice at -20oc to water at 30oc, 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 50oc, at an extra one minute. Determine latent heat of fusion/ and specific heat capacity of the solid.

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(Ѳ2- Ѳ1)

I = 10A, m =ikg, L =360000JK-1, v = 30v, t = 1200 + 0ne minute = 1200 + 60 = 1260s (Ѳ2- Ѳ1) = 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-1K-1

Specific heat capacity == 3.6 x 102JKg-1K-1

Specific latent heat = 3.6 x105JK-1

# specific heat capacity questions and answers 16

Lewis Hamilton is driving his car weighing 1000kg at the top of a hill travelling at 20m/s. He puts the transmission in neutral and using the brakes, he comes to a stop at a point 15m below its start position. If the brake discs have a mass of 5kg each how hot do they get? Assume specific heat for mild steel as 500 J/kg°C

Using

1/m1V2 = Cm2Ѳ

Where c = specific heat capacity = 500J/kg0C, m2 = mass of the brake discs = 5kg, m1 =1000kg (mass of the car) v = 20m/s (velocity of the car) and Ѳ = change in temperature.

So

½ x 1000 x 202 = 500 x 5 x Ѳ

200,00 = 2500Ѳ

Ѳ = 200,000/2500 = 800C

So Ѳ= 800C (that is how hot the disc will be)

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