This article will discuss the concept of calorimetry. We will focus on the calorimeter principle, calorimetry formula, and calorimetry problems in the following segments. Calorimetry is a very important aspect to be considered and worked upon in many industrial processes as well. It is with the help of calorimetric and thermodynamic principles we have been able to establish economic methods for industrial and production processes.
As we all know, the basic concept of calorimetry is as follows-
The heat released from a substance with higher temperature = Heat absorbed by a substance with a lower temperature.
The concept of calorimetry uses the following formula to determine the transfer of heat-
Q = m C △ T
In the above equation, Q is the unit of measure for heat transfer;
M is the mass of the substance;
C is the specific heat capacity; and
△ T is the temperature change.
Problem 1 – If the specific heat of water is 4.2 J/gm K, then what is the amount of heat needed to change the temperature of 3g of water by 60 ° C?
Solution – From the above statement, we can gather the following information-
Specific heat (C) = 4.2 J/gm K, mass (m) = 3g, and change in temperature (△ T) = 60 ° C
To this, we apply the following formula: Q = m C △ T
Q = 3 4.2 60 = 756 Joules
Therefore, the amount of heat required to change 3g of water by 60 ° C with the specific heat of water being 4.2 J/gm K is 756 Joules.
Problem 2- If the amount of heat required to change the temperature of a 0.5 kg mass of lead from 60 ° C to 100 ° C is 1,500 Joules, then determine the specific heat.
Solution – From the above statement, we can gather the following information-
mass (m) = 0.5 kg
change in temperature (△ T) = 100 ° C – 60 ° C = 40 ° C
The amount of heat required (Q) = 1500 Joules
We know that Q = m C △ T
Therefore, specific heat (C) = Qm×△ T
C = 15000.5100040 = 0.075 J/gm K ( 1kg = 1000 grams )
Therefore, the specific heat or heat capacity is 0.075 J/gm K.
Problem 3- If the amount of heat required to change the temperature of a 0.9 kg mass of lead from 35 ° C to 75 ° C is 1,800 Joules. Then, determine the specific heat.
Solution – From the above statement we can gather the following information-
mass (m) = 0.9 kg
change in temperature (△ T) = 75 ° C – 35 ° C = 40 ° C
The amount of heat required (Q) = 1,800 Joules
We know that Q = m C △ T
Therefore, specific heat (C) = Qm×△ T
C = 1,8000.9100040 = 0.05 J/gm K ( 1kg = 1000 grams )
Therefore, the specific heat or heat capacity is 0.05 J/gm K.
With this, the article on calorimetry is concluded. We hope that we were able to clarify and add to your knowledge on the calorimeter principle and calorimeter problems. Calorimetric formulas and concepts can be used to find out the heat transfer in ideal conditions. Several types of calorimeters are used based upon the requirements. As things get further complex, we introduce more thermodynamic principles to cater for those problems.