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Calorimeter is a device that is used to measure heat energy transfer or thermal energy transfer from one object to another. In brief, a calorimeter is an instrument that holds the capacity to measure calorimetry. Calorimetry follows the principle of the law of conservation of energy which implies, heat loss is equal to heat gain. In a calorimeter, two forms of matter, mainly a liquid and a solid, are situated in contact with one another, where, both bodies have distinct temperatures. This is the reason why heat energy gets transferred from an object having a greater temperature to an object having a lesser temperature. Here, we will discuss more about the calorimeter formula
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Key Terms – Calorimetry, Calorimeter, Temperature, Heat Energy, Thermal Energy, Specific Heat Capacity, Equilibrium
Also check: Potential Energy
What is Calorimeter?
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Every object is made up of atoms and molecules which are innumerable in number inside the object. When these atoms and molecules move in the specified region, they produce energy in the form of heat energy or thermal energy. Heat energy has the tendency to travel from high temperature to low temperature or from low temperature to high temperature, till equilibrium is attained. During this process, heat is measured as calorimetry. And the instrument by which calorimetry is measured is known as a calorimeter.
Calorimeter
For instance – Assume, there is ice on one side and a cup of tea on the other side. After some moment, the ice will start melting and the tea will start cooling. This condition will continue till the time, ice will convert into the water of room temperature and tea will come to room temperature. This conversion of heat from high temperature to low temperature in the condition of tea and low temperature to high, temperature in the condition ice shows the tendency of transfer of heat with respect to temperature.
This transfer of heat is measured in calorimetry and for the same, the device which is taken into consideration is known as a calorimeter.
Read More: Thermochemistry
Principle of Calorimetry
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Calorimeter is based on the principle of the Law of Conservation of Energy. It means, when an object of high temperature comes in contact with an object of low temperature, transfer of heat is attained till the time both high and low temperature objects come into equilibrium. It simply means the heat which is lost is equal to the heat which is gained.
Heat loss = Heat Gain
Principle of Calorimetry
Structure of Calorimeter
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The calorimeter is made up of a metallic vessel and a stirrer. Both apparatuses are made up of the same material, either copper or aluminum. Now, this vessel is placed inside a wooden jacket that holds a heat-insulating material such as glass, wool, etc. In this system, the outer jacket behaves as a heat shield and minimizes the heat loss from the inner vessel. In this system, there is an opening via which a mercury thermometer is placed.
Structure of Calorimeter
Read More: Application of Thermodynamics
Calorimeter Formula
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As the principle of calorimetry states, heat loss is equal to the heat gained.
So, the transfer of heat is measured as
Where,
- Q is heat evolved, (heat absorbed – heat released) in Jules (J)
- m is the mass in kilograms (Kg)
- c is the specific heat capacity in J/kg⋅°C (or J/kg⋅Δª)
- ΔT is the temperature change in °C (or Δª)
Things to Remember
- Calorimetry is the process by which the transfer of heat from one temperature to another is measured. And the device which is used to measure the calorimetry is known as a calorimeter.
- The calorimeter is based on the law of conservation of energy, which means when two objects of two different temperatures are kept in contact the heat is transferred till the time both come down to the same temperature.
- This flow of heat can be from high temperature to low temperature or from low temperature to high temperature.
- The calorimetry formula is Q = mC (delta) T.
Sample Questions
Ques: What is the amount of heat needed to change 1g of water by 20°C. Provided that C of water is 4.2 J/gm K. (2 Marks)
Ans: In the given question,
C= 4.2 J/gm K;
m= 1g;
ΔT= 20,
Q =?
As per the formula of calorimeter, the equation is
Q= mCΔT
Therefore, Q= 1 X 4.2 X 20 =
Q = 84 Joules
Ques: The amount of heat needed to change 50 g of water by 40°C. The specific heat capacity of water is 4.2 j/gm K. Find the heat evolved. (2 Marks)
Ans: In the given question,
Q =?
ΔT= 40
m = 50
C = 4.2 j/gm K
Therefore, as per equation of calorimeter, Q= mCΔT
Q = 50 x 4.2 x 50
Q = 8400 Joules
Ques: A piece of ice of mass 40g is added to 200g of water at 50°C. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water=4200kg k and specific latent heat of fusion of ice=336 J Kg. (4 Marks)
Ans: In the given question,
Let t ºC is the final temperature
m (ice) = 40 g
m (water) = 200 g
C (ice) = 336
C (water) = 4200 kg
Q (ice) = 40×336+40×t
Q (water) = 200×4.2×(50-t)
As per, principle of calorimeter, heat loss = heat gain
Q (ice) = Q (water)
40×336+40×t = 200×4.2×(50-t)
t = 32.5 ºC
Ques: Find the result of mixing 10 g of ice at -10οC with 10 g of water with 10οC. Specific heat capacity of ice=2.1 J g-1 K-1, specific latent heat of ice=336 J g-1, and specific heat capacity of water=4.2 J g-1K-1. (4 Marks)
Ans: In the given question,
As per the principle of calorimeter some ice will melt and become water and the whole mixture will reach a temperature of 0ºC.
let x gm of ice will melt.
heat energy gained by ice Q (ice) = (x×2.1×10 + x×336) J
heat energy lost by water Q (water) = 10×4.2×10 = 420 J
As per the principle,
Heat loss of ice equals heat gain of water,
x × (21+336) = x × 357 = 420
So, we get x = 1.176 g
Hence,
The final mixture is 11.176 g of water + 8.824 g of ice and the mixture is at the temperature of 0ºC.
Ques: A solid of mass 50g at 150°C is placed in 100g of water at 11°C, when the final temperature recorded is 20°C. Find the specific heat capacity of the solid. (Specific heat capacity of water= 4.2Jgâ»1oCâ»1. (4 Marks)
Ans: In the given question,
Q (solid), Heat lost by solid = ms×S×ΔT = 50×10-3×S× (150-20) = 6500×10-3×S Joules
Where,
mS is mass of solid
S is specific heat
and ΔT is change in temperature
Q (water) Heat gained by water = mw×Sw×ΔT = 100×10-3×4200× (20-11) = 3780 Joules
Where,
mw is the mass of water
Sw is specific heat of water
By the principle, Heat loss = Heat Gain,
So,
6500×10-3×S = 3780;
So,
S = 582 kJ/ (kg K)
Ques: Calculate the mass of ice needed to cool 150g of water contained in a calorimeter of mass 50g at 32 degrees C such that the final temperature is 5-degree C. Specific heat capacity of calorimeter = 0.4 j/g degrees C. Specific heat capacity of water = 4.2 j/g degree C and Latent heat capacity of ice = 330j/g. (4 Marks)
Ans: In the given question as pe the principle,
Heat loss by (water + Calorimeter) = Heat gain by ice
Heat loss by (water + calorimeter) = mw Cpw ΔT + mC Cpc ΔT = mi (L + Cpw δT )
Where,
mw = mass of water = 50 g
Cpw = Specific heat of water = 4.2 J/ (g °C)
mC = mass of calorimeter = 50 g
Cpc = Specific heat capacity of calorimeter = 0.4 J/ (g °C)
ΔT = fall in temperature of water and calorimeter = 32-5 = 27°C
mi = mass of ice in gram
L = latent heat capacity of ice = 330 J/g
δT = rise in temperature = 5 °C
by substituting all the values
we get mass of ice,
g
Ques: Suppose the masses of calorimeter, the water in it and the hot object made up of copper which is put in the calorimeter are the same. The initial temperature of the calorimeter and water is 30°C and that of the hot object is 60°C. The specific heat of copper and water are 0.09 cal/gm°C and 1 cal/gm°C respectively. Also, the specific heat of the calorimeter is the same as copper. What will be the final temperature of water (mixture). (4 Marks)
Ans: In the given question,
Heat gain by water and calorimeter = heat gain by hot object.
let m (in gram) be the mass of each i.e., water, calorimeter, and hot object.
calorimeter and hot object both are made up of copper.
Let T be the final temperature.
Heat gain by water and calorimeter = m×0.09×(T-30) + m×1×(T-30) .......................1
Heat loss by hot object = m×0.09×(60-T) ....................................... 2
By equating (1) and (2), we get, (T-30) ×1.09 = (60-T) ×0.09 ...............3
solving for T from eqn. (3),
we get T = 32.3 °C
Ques: An unknown metal of mass 192 g heated to a temperature of 100ºC was immersed into a brass calorimeter of mass 128 g containing 240 g of water a temperature of 8.4ºC Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5ºC (Specific heat of brass is 394 J kg–1 K–1) (1) 1232 J kg–1 K–1 (2) 458 J kg–1 K–1 (3) 654 J kg–1 K–1 (4) 916 J kg–1 K–1. (4 Marks)
Ans: In the given question,
Water equivalent = (0.128×394)/4200 =
12.00gms
Mw × Sw = Mmet × Smet
Mw = (Mmet × Smet)/Sw
Total water = 240 + 12 = 252 gm
Hence,
Heat lost by metal= 192 × Sm × (100 – 21.5) = 15072
Heat gained by warter= 252 × 4200 × (21.5 – 8.4) = 13865040
So,
Sm = 916 J/kgK
Ques: 104 g of water at 30°C is taken in a calorimeter made of copper of mass 42g. When a certain mass of ice at 0°C is added to it, the final steady temperature of the mixture after the ice has melted, was found to be 10°C. Find the mass of ice added. (4 Marks)
Ans: In the given question,
Heat loss of (water + Copper calorimeter) = ( mw Cpw + mC Cpc ) ΔT ......................1
Where,
mw = mass of water = 104× 10-3 kg
Cpw = Sp. Heat of water = 4200 J/ (kg oC)
mC = mass of calorimeter = 42×10-3 kg
Cpc = Sp.Heat of copper = 385 J/ (kg oC)
ΔT = temperature difference = 30 - 10 = 20 oC
Now,
Let m be the mass of ice added to water
Heat gain by ice = m×(L+Cpw × ΔT) = m × (334×103+ 4200 × 10) ......................2
Where,
L is latent heat of fusion of ice,
L = 334 kJ/kg
Now,
By substituting all the values in eqn.(1) and considering heat loss of water+Calorimeter is equal to heat gain by ice,
Solving for mass of ice m to get
m = [ (104 × 10-3 × 4200 + 42 ×10-3 × 385) × 20] / (334×103 + 42000)
= 0.024 kg =
m = 24 g
Ques: 1.5 kg of ice melts when a jet of steam at 100 degrees C is passed in an ice block. Calculate the mass of steam required to achieve it. (4 Marks)
Ans: In the given question, Let,
MS be the required mass of steam in kg.
Heat loss by the steam from 100°C to oC is,
QL = MS×Lv+MS×Cp×(100-0)
Where,
Lv is latent heat of vaporization of steam,
Cp is specific heat of water
Heat required to Melt 1.5 kg of ice,
QG = 1.5×Lf
Where Lf is latent heat of vaporization of steam
QL = QG
MS×Lv+MS×Cp×(100-0) = 1.5×Lf
MS = 1.5 × [ Lf / ( Lv + 100 × Cp)]
Lf = 334 kJ/kg,
Lv = 2260 kJ/kg
Cp = 4.2 kJ/(kgK)
So, MS = (1.5×334) /2680
MS = 187 gm.
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