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Quality of heat Questions

1.  (a) Define the term specific heat capacity.  

(b) A block of metal of mass 300g at 1000c is dropped into a logged calorimeter of heat capacity

40Jk-1, containing 200g of water at 200c. The temperature of the resulting mixture is 340c.

(Specific heat capacity of water = 4200Jkg-1k-1)

Determine:

(i) Heat gained by calorimeter.

(ii) Heat gained by water.

(iii) Heat lost by the metal block.

(iv) Specific heat capacity of the metal block.

 

 

 

2.  (a) State two differences between boiling and evaporation.

(b) 200g of a solid was uniformly heated by a 0.2 kw heater for sometime. The graph in the

figure below shows how the temperature of the solid changed with time.

ecolebooks.com

 

(i) Explain what is happening between OA and AB.

 (ii) Calculate the specific heat capacity of the solid.

(iii) Calculate the specific latent heat of fusion k of the solid.  

 

3.  (a) Define the term heat capacity    

 (b) A block of metal of mass 150g at 100oC is dropped into a logged calorimeter of heat capacity

  40Jk-1 containing 100g of water at 25oC. The temperature of the resulting mixture is 34oC.

(Specific heat capacity of water = 4200J/KgK)

 Determine;-

(i) Heat gained by calorimeter

(ii) Heat gained by water

(iii) Heat lost by the metal block  

(iv) Specific heat capacity of the metal block

 

4.  (a) Distinguish between evaporation and boiling

 (b) A jet delivering 0.44g of dry steam per second, at 100oC is directed on to crushed ice

  at 0.0oC contained in an unlagged copper can which has a hole in the base. 4.44g of water

at 0.0oC flow out of the hole per second

(i) How many joules of heat are given out per second by condensing steam and cooling to

0.0oC of water formed?(Latent heat of vaporization of steam = 2.26 x 106JKg-1,

c for water = 4200JKg-1K-1)

(ii) How much heat is taken in per second by the ice which melts?  

(iii) Suggest why these amounts above are different

Image From EcoleBooks.com  (c) Figure 7 below shows a cross-section of a vacuum flask

 

 

 

 

 

 

 

 

 

 

 

 

  (i) Name the parts labelled A and B on the diagram

 (ii) Explain how the heat losses are minimized when hot liquid is poured into the flask  

 

5.  (a) Figure 2 shows two identical thermometers. Thermometer A has a blackened bulb while

thermometer B has a silvery bulb. A candle is placed equidistant between the two thermometers

 

 

 

 

 

 

Image From EcoleBooks.com

 

 

 

 

 

 

 

 State with a reason the observations made after some time

5.  (b) Figure 3 shows a test tube partially filled with water. An ice wrapped in wire gauze is

Image From EcoleBooks.com   placed at the bottom of the test-tube. It is then held in the flame of a bunsen burner as

shown below

 


fig. 3

 

 

 

 

 State and explain what will be observed after some time

6.  Give any two differences between evaporation and boiling

7.  Explain why steel feels colder than wood at the same temperature

 

8.  An electric heater 1KW 240V is used to raise the temperature of a 5kg copper block from

15oC to 33oC. If the specific heat capacity of copper is 400JKg-1K-1 and assuming no heat is

lost to the surrounding,

 Calculate the time taken

 

9.  (a) Define specific latent heat of fusion

(b) 0.5kg of naphthalene contained in an aluminium can of mass 0.4kg is melted in a water

  bath and raised to a temperature of 100oC . Calculate the total heat given out when the

  can and its contents are allowed to cool to room temperature, 20oC . Neglect losses by  

  evaporation during heating process and give your answer to the nearest kilojoule.

 (For naphthalene melting point = 80oC , Specific heat capacity for both liquid and

  solid =2100J/KgK; specific latent heat of fusion = 170000J/Kg.

  For aluminium: specific heat capacity = 900J/Kgk  

 (c) Briefly explain two ways other than direct heating by which quantity of liquid may be made to

  evaporate more quickly

(d) The diagram below shows a charcoal refrigerator

 

 

 

 

 

Image From EcoleBooks.com

 

 

 

 

 

 

 

 

 (i) Explain why charcoal is used and why it is sprinkled with water  

 (ii) What is the role of the metallic tank and the wire mesh

 

10.  An electric kettle with a shinny outer surface would be more efficient than one with a dull

outer surface. Give a reason for this

 

11.  A heating element rated 2.5 KM is used to raise the temperature of 3.0 kg of water through

  50oC. Calculate the time required to Effect this. (Specific heat capacity of water is 4200J/kgK).

 

 

Quality of heat Answers

1.  a) Specific heat capacity is defined as the quantity of heat required to raise the temperature of

a unit mass of a substance by one Kelvin. 1 mk

b) (i) heat gained by calorimeter

QE = C∆θ

= 400Jk-1 x (34 – 20)k 1   = 560j 1

 

  (ii) Heat gained by water

QW= MC∆θ = 200 kg x 4200kg-1k-1 x(34 – 20)k

  = 11760j 1

  (iii) Heat lost by the metal = heat gained by the calorimeter + heat gained by Water.

= 560j + 11760j = 12320j 1

  (iv) Specific heat capacity of the metal.

   C = Q
1

  M∆θ

  = 12320 j
1

  0.3 kg x (100 – 34)

 = 12320j = 622.22 jkg-1k-1
1

  0.3kg x 66 k

2. (a)

Boiling

Evaporation

  • Takes place at constant temperature
  • Not affected by surface area of liquid exposure
  • Vigorous, visible process al over the liquid
  • Takes place at all temperatures
  • Increases with increase in surface area
  • Slow invisible process at the liquid surface

(each for any two correct)

(b) (i) A – Temperature of the solid in increasing;

AB- solid is melting;

 

(ii) Heat supplied = Heat gained

  p x t = MC

  200 x 100 = 0.2 x C x 100

  20 x 100 = 0.2 x C x (350 – 250)

C= 1000JKg-1 K-1;

 

(iii) Heat applied = heat gained; OR

p x t = Mlf;

200 x 200 = 0.2 x lf;

lf = 200,000J/Kg;

3.  (a) The quantity of heat required to raise the temperature of a given mass of a material

by one Kelvin.

  (b) (i)   C = Q



Q = C

= 40 x (34-25)

= 40 x 9 = 360J

(ii) MwCwD = (0.1 x 9 x 4200) = 3780J

 

(iii) Heat lost = heat gained by calorimeter + heat gained by water

= 3780 + 360 = 4140J or 4.14KJ

(b) 150 x Cm x  = 4140

1000

0.15 x (100 – 34) Cm = 4140

9.9C = 4140

Cm = 4140 = 48.18JKg-1K-1

9.9

(b)

Boiling

Evaporation

  • Temperature is constant
  • Affected by impurities and pressure
  • vigorous with bubbles
  • takes place in the whole liquid
  • Temperature can vary
  • Not affected with impurities
  • not vigorous, no bubbles
  • takes place at the surface

4.  (a) Evaporation is change of liquid to gas at any temperature while boiling is change of liquid

to gas at constant temperature;

(i) Heat – mlv + MC

= 0.44 x 2.26 x 106 + 0.44 x 4200 x 100;

1000 1000

=994.4 + 184.8 = 1179.2J;

(ii)  Q = mLf

  t t

= 4.4 x 10-3 x 3.34 x 105

= 1336J;

(iii)  – Heat is absorbed from the surrounding since the can is unlagged;

(c)  A = Vacuum;

B = Silvered (shiny) surfaces/polished/smooth;

(ii)  – Loss of heat by conduction is reduced by cork and vacuum;

– Loss of convection is reduced by vacuum;

– Loss by radiation is reduced by silvered wall;

5.  Thermometer A gives a higher reading than B;  1 Black surface is better absorber of heat  than silvery surface1

5. Water at the top part of the boiling tube boils while the ice remains unmelted. This is because

water is a poor conductor of heat and hot air less dense remain at the top.

6.   – Evaporation is silent while boiling is a vigorous visible process

  – Evaporation takes place at all temperatures while boiling takes place at a specific

temperature

– Evaporation takes place on the surface of the liquid but boiling takes place in the whole

liquid

7.  Steel is a good conductor of heat; therefore draws heat from your body unlike wood

which is a poor conductor of heat

8.  Pt = MCDT  DT = 35 – 15 = 18K

T = MCDT

= 5kg x 400JKg-1K-1 18K

1000JS-1

= 36s

 

9. (a) It is the quantity of heat required to convert a unit mass of the substance from the solid to  the liquid state without change of temperature  (1mk)

Heat lost by naphthalene

T = 100-80=20k

H1 = 0.5kg x 2100JKg-1K-1 x 20K = 2100J

Lf = mLf = 0.5kg x 170000JKg-1 = 85000J

T = 80 -20 = 60k , H2 = 0.5kg x 2100JKg-1K-1 x 60k = 63000J

Heat lost by aluminium

T = 100 – 20 = 80k

H = 0.4kg x 900JKg-1K-1 x80k = 28800J

Total heat lost  = 169000J + 28800J = 197800J

= 197.8KJ

= 198KJ

(c) Blowing wind over the surface of the liquid increases the kinetic energy of the

liquid molecules

– It can also be made to evaporate faster by bubbling air though the liquid as it increases the

surface area from which the liquid molecules may escape

(d) (i) – Charcoal is a black body and therefore a batter absorber of heat

– Sprinkled with water, so that it takes latent heat from the air around the air around

and evaporates , causing cooling in cabinet

(ii) Heat is conducted by the by the metallic tank and the wire mesh to the surrounding air

10.  Loss of heat through radiation is reduced/ minimized


 




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