INTRODUCTION: Heat is a kind of energy that make a substance colder or hotter while temperature is the degree of hotness or coldness of a substance. ?(calorimetry)

DEFINITION OF CALORIMETRY:  It is a science of measurement which deals with heat and temperature. We know that one calorie of heat energy is defined as the amount of heat required to raise the temperature of 1 m of water from 14.5°C to 15.5°C.( calorimetry)

PRINCIPLES OF CALORIMETRY:  In thermal equilibrium of two bodies heat given by hot body is equal to heat loss by cold body (calorimetry).

Heat taken by the hot body = Heat taken by the cold body (calorimetry)

ESTABLISHMENT OF CALORIMETRY FORMULA: (calorimetry) Let two bodies of masses m1 and m2 with specific heat c1and c2 are  at the temperature of Ɵ1 and Ɵ2 respectively.  (calorimetry)Let temperature of the first body is more than that of second body (Ɵ1 > Ɵ2) (calorimetry). Now two bodies are put in contact. (calorimetry) As first body being hotter transfers the heat energy to the second body. ( calorimetry) This transfer of heat energy continues till the two bodies are coming at common temperature Ɵ. (calorimetry)

Let’s calculate the fall in temperature of the first body (calorimetry) = Ɵ1 – Ɵ

Rise in temperature of the second body (calorimetry) = Ɵ –  Ɵ2

Amount of heat given by the first body  (calorimetry) = m1 X c1 X (Ɵ1 – Ɵ) (calorimetry)

Amount of heat taken by the second body (calorimetry) = m2 X c2 X (Ɵ  – Ɵ2) (calorimetry)

In thermal equilibrium when there is no loss of heat energy, (calorimetry)

Then, m1 X c1 X (Ɵ1 – Ɵ ) =m2 X c2 X (Ɵ – Ɵ2) (calorimetry)

ISOTHERMAL TITRATION CALORIMETRY: It is a physical technique (isothermal titration calorimetry) used to determine the thermodynamic parameters of interactions in solution (isothermal titration calorimetry). It is the most often used  process to study the binding of small molecules to larger macromolecules (isothermal titration calorimetry). It consists of two cells (isothermal titration calorimetry) which are enclosed in an adiabatic jacket.(isothermal titration calorimetry)

CONE CALORIMETER:  A cone calorimeter is a device used  to study the fire behavior of small sample of various material  (cone calorimeter) in condensed phase (calorimetry).  A cone calorimeter is widely used in the field of fire safety engineering (cone calorimetry).

In mid of 80’s there was a need of measuring material flammability based on rate of heat release.(cone calorimeter). It is believed that heat releasing rate is the most authentic and accurate measure of flammability of a material (cone calorimeter). That time, very few  heat release measuring tools were available which were not reliable, difficult to handle and inconsistent data. So, in 1982 NIST introduced the cone calorimeter to measure material flammability. (cone calorimetr)In mid 80, cone calorimeter was commercially available. Now, there are more than 300 cone calorimeter  found world-wide. ASTEM E1354 and D5485, ISO5660-1, NFPA271 and CAN\ULC- S135 are the best cone calorimeter.


Qs1 (calorimetry):   100gm of hot water at 80°C is mixed with 200 gm of cold water at10°C. what is the temperature of the mixture?(calorimetry)

Ans1 (calorimetry) : Given

 m1= 100gm

m2 = 200gm

c1 = c2 = 1.0 calorie /gm°C

Ɵ1= 80°C

Ɵ2 = 10°C

Therefore, according to law of calorimetry

  So, m1 X c1 X ( Ɵ1 – Ɵ ) = m2 X c2X (Ɵ – Ɵ2)

  • 100 X 1 X (80 – Ɵ) = 200 X 1 (Ɵ – 10)
  • 80 – Ɵ = 2 X(Ɵ -10)
  • Ɵ = 33.3°C
  • So the required temperature is 33.3°C. (calorimetry)

Qs2 (calorimetry) A piece of metal of mass 250 gm is heated to the temperature 300°C (calorimetry). Hot metal is put in 120gm of water at 20°C contained in a calorimeter of mass 200gm . (calorimeter) Final temperature of mixture is 40°C . (calorimeter) Calculate the specific heat capacity of metal (calorimeter). Specific heat of calorimeter = 0.1 cal/gm°C. (calorimeter)

Ans2 (calorimetry) :  heat given by the solid Q = mcƟ

So, Q1= 250  X c X (300 – 40) = 65000c cal (calorimetry)

Q2 = 120 X 1 X (40 – 20) = 2400 cal (calorimetry)

Q3 = 200 X 0.1 X (40 – 20) = 400 cal (calorimetry)

Q1 = Q2 + Q3

=> 65000c = 2400 + 400

è c = 2800/ 65000 = 0.04cal /gm°c (calorimetry)

Qs3 (calorimetry) A copper calorimeter of mass 200gm contains 150gm of water at 10°C.( calorimetry) A piece of metal of mass 400gm heated upto temperature 250°C is put in the calorimeter. The final temperature of mixture becomes 30°C. What is the specific heat of metals?(calorimetry ) Specific heat of calorimeter = 0.1cal/gm°C.(calorimetry)

Ans3 (calorimetry) : Heat given by metal = 400 X c X (250 – 30) = 88000c (calorimetry)

Heat taken by water = 150 X 1 X (30 -10) = 3000 calorie (calorimetry)

Heat taken by calorimeter = 200 X 0.1 X (30 – 10) (calorimetry)

Heat given  by metals = heat taken by water + heat taken by calorimeter

88000 X c = 3000 + 400 (calorimetry)

  • C = 3400/88000 (calorimetry)
  • Specific heat of metal = 0.04cal/ gm°C (calorimetry)

Qs4 (calorimetry) : 400gm liquid 9specific heat = 0.5 cal/ gm°C  at 30° C is put in a calorimeter of 250 gm. An iron piece of mass 350gm heated to temperature 200°C is put in this calorimeter.  (calorimetry)What is the final temperature of the mixture .  (specific  heat of iron = 0.12 cal /gm° C ; specific heat of calorimeter = 0.1 cal/gm°C.(calorimetry) .

Ans4 (calorimetry): Let final temperature of mixture is Ɵ

Heat g =

iven by iron = 350 X 0.12 X (200 –Ɵ ) = 42(200 –  Ɵ) (calorimetry)Ɵ

Heat taken by the liqid = 400 X  0.5 X (Ɵ –30) (calorimetry)= 200( Ɵ – 30)

Heat taken by calorimeter  = 250 X 0.1  x ( Ɵ -30) = 25 X (Ɵ -30)

According  to calorimetry ,

42X (200 – Ɵ) = 200 X ( Ɵ – 30) + 25 X (Ɵ -30) (calorimetry)

  • 42 X (200 –Ɵ ) = 225 X (Ɵ – 30)
  • 42 X 200 – 42Ɵ = 225Ɵ -225 X 30
  • 225Ɵ + 42Ɵ = 42 X 200 + 225 X 30
  • Ɵ = 51°C (approx) (calorimetry).

Qs5 (calorimetry) : 10gm of ice at 0° C is converted to steam at 100° C . calculate the amount of heat given in the process (calorimetry)

Ans5 (calorimstery) Heat given to melt ice  to come to water at 0°C = m X L = 10  X 80 = 800 = 800 cal. (calorimetry)

Heat given to water  at 0°C to come to at 100 °C (calorimetry)  = 10 X 1 X (100 – 0)  = 1000cal (calorimetry)

Heat given to  water at 100°C to convert to stem at  same temperature = 10 X 540 = 5400cal

Amount of total heat given = 800 + 1000 + 5400 = 7200 cal (calorimetry)

Therefore, calorimetric solution is 7200 cal (calorimetry)

Therefore,  calorimetric solution is 7200 calorie.  (calorimetry)

Therefore, calorimetric solution is 7200 calorie. Total heat will be required in the process. (calorimetry).  




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