# Thermodynamics Equation Calculator

In this Physics tutorial, you will learn:.

How does the process of heat transfer occur? What does the general law of calorimetry say? Which are the four methods of heat transfer? What is a calorimeter used for? How can we calculate the rate of heat transfer in each of the four methods of heat transfer? What does a blacksmith do when he wants to cool down a piece of hot processed iron?

Does he wait until the iron cool down by itself or he uses any fast method for this?

What do you obtain if mix hot water and cold water? Is it relevant the amount of each water sample?

Suggest a method to measure the heat supplied to a substance.

How many methods of heat transfer do you know? What do they have in common? This tutorial will focus on the measurement of heat exchange between thermodynamic systems.

In scientific terminology, this process is known as "calorimetry".

Also, the methods of heat transferred and their calculation will be extensively discussed. Let's have a closer look at all these.

As explained in the previous tutorial "Absorption of Heat.

States of Matter. Change of State." , when two substances are placed in contact, there is a heat flow between them.

The direction of heat flow is pre-determined; heat always flows from the hottest to the coldest object.

This "heat flow" does not necessarily imply any matter transfer; rather, it often takes place when the boundary atoms of the two objects collide with each other at different speeds.

Such a collision only makes the slow vibrating molecules of the cold object vibrate faster due to their collision with the fast vibrating molecules of the hot object, as discussed in the previous tutorials.

However, in many cases there is also some matter transfer during the process of heat exchange between systems.

This occurs for example when we mix some hot and cold water.

The result will be an amount of warm water, whose mass is the sum of the individual masses. Obviously, it is expected that the warm water have an in-between temperature. The general law of calorimetry states that: . During a heat exchange process between two objects of a thermodynamic system, the heat released by the hottest object is entirely gained by the coldest object if the system is isolated from the surroundings. Mathematically, we can write . Qreleased by the hot object = Qgained by the cold object. We can write for the heat released by the hottest object.

where m1 is the mass of the hottest object, c1 is its specific heat capacity, t1 is the initial temperature of the hottest object and tf is the final (common) temperature after the heat exchange process is completed.

On the other hand, we can write for heat gained by the coldest object. Q2 = m2 × c2 × ∆t2 = m2 × c2 × (tf - t2 ).

where m2 is the mass of the coldest object, c2 is its specific heat capacity, t2 is the initial temperature of the coldest object and tf is the final (common) temperature after the heat exchange process is done.

Pay attention to the expressions within the parenthesis. In the expression for the hottest object, we subtract the final temperature after heat exchange from the initial temperature as it gives a positive value, while in the expression for the coldest object we subtract the initial temperature before the contact from the final temperature in order to obtain a positive value.

Therefore, combining the two above equations, we obtain the four equivalent equations (which basically represent the same thing): . Qreleased by the hot object = Qgained by the cold object. m1 × c1 × ∆t1 = m2 × c2 × ∆t2. 20 g water at 80°C is mixed with 30 alcohol at 10°C.

Find the final temperature if no heat exchange with the surroundings does occur. Take the specific heat capacity of water as 1 cal/g°C and that of alcohol as 0.58 cal/g°C.

Using the notations introduced earlier in the theory, we write the following clues: .

m1 = 20 g m2 = 30 g t1 = 80°C t2 = 10°C c1 = 1 cal/g°C c2 = 0.58 cal/g°C tf = ? Applying the equation of heat exchange .

m1 × c1 × (t1 - tf ) = m2 × c2 × (tf - t2 ). we obtain after substituting the known values: . 20 × 1 × (80 - tf ) = 30 × 0.58 × (tf - 10) 20 × (80 - tf ) = 17.4 × (tf - 10) 1600 - 20 × tf = 17.4 × tf - 174 37.4 × tf = 1774 tf = 1774/37.4 =47.4°C.