Find Thermal Equilibrium: 7 Important Facts You Should Know

In this article, the topic, “find thermal equilibrium” will be summarizing in a brief manner. In the state of thermal equilibrium the heat is no transfer in between two substances which are situated in contact.

From the zeroth law of the thermodynamics we get a clear concept about the thermal equilibrium. Thermal equilibrium is a condition in between two objects where the heats in not transfer and objects have contact with each other. The temperature for the two objects remains same for thermal equilibrium condition.

A very interesting conception related to temperature the term is thermal equilibrium. Two substances are in the state of thermal equilibrium if in a close system the temperature is gained or rejected by the objects until they reached to the state of equilibrium in spite of that, energy is not transfer in between the two substances. As same as, when the substances are not in contact they also stay at thermal equilibrium state if when situated in contact nevertheless the energy is not exchange in between two substances.

Examples of thermal equilibrium:-

Some examples of the thermal equilibrium is discuss below,

1. Measuring the temperature of human body by thermometer is a nice example for thermal equilibrium. When temperature is examine by the help of thermometer the temperature is gain by the thermometer and after a certain time when the temperature of the body and thermometer became same heat transfer in between body and thermometer became stop, means the state reaches to equilibrium condition.
2. Suddenly put hand into ice cube is another nice example of thermal equilibrium. When hand put into ice cube at that time temperature in between cube and hand start to transfer the heat and when the temperature in between hand and ice cube became same the heat transfer will be stop, means the state reaches to equilibrium condition.
3. Melting of a butter cube is also example of thermal equilibrium. When a cube of butter is placed in natural at that time temperature in between cube and atmosphere try to reaches at same temperature thus start to transfer the heat, and when the temperature in between atmosphere and cube of butter became same the heat transfer will be stop, means the state reaches to the equilibrium condition.

How to find Thermal equilibrium?

The thermal equilibrium is a state when heat transfer of two substances will be stop when their temperature reaches to same point. The thermal equilibrium is determine by the help of this formula,

$Q = m \times C_e \times \Delta T$ …eqn(1)

Where,

Q = Total energy of the specific matter of the body which is expressed in Joule

m = Mass of the specific matter of the body which is expressed in grams

$C_e$ = Specific heat of the specific matter of the body which is expressed in joule per Kelvin per kilogram

$\Delta t$ = Temperature of the specific matter of the body which is expressed in Kelvin.

If the value of mass of a substance and specific heat, temperature is given then easily putting the values in …eqn(1) we can determine the value of thermal equilibrium.

Now with a help of numerical problem the thermal equilibrium is discuss,

Problem:

For the 40 gram of water to reaches the temperature for 45 degree centigrade. Now determine the value of energy at which the water stays at thermal equilibrium state.

Solution:-

In the problem given data are,

Mass of the water (m) = 40 gram

Specific heat for the water is $(C_e)$ = 4.17 Joule per gram degree centigrade

Temperature $(\Delta T)$ = 45 degree centigrade

We know that, the formula for thermal equilibrium is,

$Q = m \times C_e \times \Delta T$

Where,

Q = Total energy of the specific matter of the body

m = Mass of the specific matter of the body

$C_e$ = Specific heat of the specific matter of the body

$\Delta T$ = Temperature of the specific matter of the body

Now we putting the values in the equation,

$Q = 40 \times 4.17 \times 45$

Q = 7506 Joule.

For the 40 gram of water to reaches the temperature for 45 degree centigrade. The value of energy at which the water stays at thermal equilibrium state is 7506 Joule.

When to find Thermal equilibrium?

When two objects are placed in contact heat (energy) is transferred from one to the other until they reach the same temperature (are in thermal equilibrium). When the objects are at the same temperature there is no heat transfer.

Formula to calculate thermal equilibrium:

Heat is the flow of energy from a high temperature to a low temperature. When these temperatures balance out, heat stops flowing, then the system (or set of systems) is said to be in thermal equilibrium.

The formula for thermal equilibrium is,

$Q = m \times C_e \times \Delta T$

Where,

Q = Total energy of the specific matter of the body which is expressed in Joule

m = Mass of the specific matter of the body which is expressed in grams

$C_e$ = Specific heat of the specific matter of the body which is expressed in joule per Kelvin per kilogram

$\Delta t$ = Temperature of the specific matter of the body which is expressed in Kelvin.

How do you find the final temperature in thermal equilibrium?

With the help of numerical the final temperature in thermal equilibrium is describe in below,

An iron piece which mass is 220 gram. The temperature of the iron piece is 310 degree centigrade. Suppose the iron piece is drooped at a container which is filled with water. The weight of the water is 1.2 kg and temperature will be 22 degree centigrade.

Determine the final temperature in thermal equilibrium for water.

Solution:-

Let, the final temperature in thermal equilibrium for water = T degree centigrade.

Now change in temperature,

$\Delta T = (T_f_i_n_a_l – T_i_n_i_t_i_a_l)$

Change in heat,

$Q = m \times c \times \Delta T$

So, change in heat of piece of the iron is,

$\Delta Q_i_r_o_n = \frac{220}{1000}\times 450 \times (T – 310) J$

$\Delta Q_i_r_o_n = 99 (T – 310) J$

Now, change in heat of water is,

$\Delta Q_w_a_t_e_r = 1.20 \times 4200 \times (T – 22) J$

$\Delta Q_w_a_t_e_r = 5040 (T – 22) J$

Using law of conservation of energy we can write,

$\Delta Q_i_r_o_n + \Delta Q_w_a_t_e_r$ = 0

Putting the values we get,

99 (T – 310) + 5040 (T – 22) = 0

99 T – 30690 + 5040T – 110880 = 0

5138 T = 141570

$T = \frac{141570}{5138}$

An iron piece which mass is 220 gram. The temperature of the iron piece is 310 degree centigrade. Suppose the iron piece is drooped at a container which is filled with water. The weight of the water is 1.2 kg and temperature will be 22 degree centigrade.

The final temperature in thermal equilibrium for water is 27.5 degree centigrade.

Problem: 1

How much energy is required to rise up the temperature for 55 degree centigrade for the 40 gram of water?

Solution:-

Given data are,

Mass of the water (m) = 40 gram

Specific heat for the water is $(C_e)$ = 4.17 Joule per gram degree centigrade

Temperature $(\Delta T)$ = 55 degree centigrade

We know that, the formula for thermal equilibrium is,

$Q = m \times C_e \times \Delta T$

Where,

Q = Total energy of the specific matter of the body

m = Mass of the specific matter of the body

$C_e$ = Specific heat of the specific matter of the body

$\Delta T$ = Temperature of the specific matter of the body

$Q = 40 \times 4.17 \times 55$

Q = 9174 Joule.

9174 Joule energy is required to rise up the temperature for 55 degree centigrade for the 40 gram of water.

Problem: 2

Ramesh who is friend of Ratan have a hobby of collecting stones. While Ramesh collect stones always he stones them in container. The container is made with metal aluminium. The weight of the container is 15.2 gram. Naturally the temperature for the container is about 36 degree centigrade. Now Ramesh in the aluminium container cold water pour. The temperature of the water will be 22 degree centigrade and weight of the water is 32 gram.

Determine the exact temperature at which, the temperature of the aluminium container and the temperature of the cold water will be same.

Solution: –

We know that, the formula for thermal equilibrium is,

$Q = m * C_e * \Delta t$

Where,

Q = Total energy of the specific matter of the body

m = Mass of the specific matter of the body

$C_e$ = Specific heat of the specific matter of the body

$\Delta t$ = (Final temperature – Starting temperature) of the specific matter of the body

For aluminium,

$Q_A = m_A * C_e_A * \Delta t_A$ ………….. eqn (1)

Given data are,

$m_A$ =  15.2 gram

$C_e_A$  = 0.215 calorie per gram degree centigrade

$\Delta t_A$ = $(T_f – T_i_A)$ degree centigrade = $(T_f – 36)$ degree centigrade

For water,

$Q_W = m_W * C_e_W * \Delta t_W$ ………….. eqn (1)

Given data are,

$m_W$ = 32 gram

$C_e_W$ = 1 calorie per gram degree centigrade

$\Delta t_W = (T_f – T_i_W)$ degree centigrade = $(T_f – 22)$ degree centigrade

Now, from………….. eqn (1) and ………….. eqn (2) we can write,

$Q_A = m_A * C_e_A * \Delta t_A = (-) Q_W = (-) m_W * C_e_W * \Delta t_W$

Putting the value from eqn (1) and eqn (2),

$15.2 \times (0.215) \times (T_f – 36) = (-) 32 \times 1 \times (T_f – 22)$

(Put the value for $C_e_W$ = 1 calorie per gram degree centigrade

$3.268 \times (T_f – 36) = -32 (T_f – 22)$

$3.268 T_f – 117.648 = -32 T_f + 704$

$3.268 T_f + 32 T_f = 704 + 117.648$

$T_f = \frac{704 + 117.648}{35.268}$

$T_f$ = 23.2 degree centigrade

Ramesh who is friend of Ratan have a hobby of collecting stones. While Ramesh collect stones always he stones them in container. The container is made with metal aluminium. The weight of the container is 15.2 gram. Naturally the temperature for the container is about 36 degree centigrade. Now Ramesh in the aluminium container cold water pour. The temperature of the water will be 22 degree centigrade and weight of the water is 32 gram.

The exact temperature at which, the temperature of the aluminium container and the temperature of the cold water will be same is 23.2 degree centigrade.

Problem: 3

An unspecified metal is store into the laboratory.  The weight of the unspecified metal is 6 gram. Now 248.2 Joule energy is added into the unspecified metal. The temperature of the unspecified metal is rise up to 116 degree centigrade.

Now, determine the amount of specific heat for the unspecified metal.

Solution:-

Given data are,

Mass of the unspecified metal (m) = 6 gram

Need to calculate,

Specific heat for unspecified metal $(C_e)$ =? Joule per gram degree centigrade

Temperature $(\Delta T)$ = 116 degree centigrade

Total energy of the unspecified metal (Q) = 248.2 Joule

We know that, the formula for thermal equilibrium is,

$Q = m \times C_e \times \Delta T$

$C_e = \frac{Q}{m\Delta T}$

Where,

Q = Total energy of the specific matter of the body

m = Mass of the specific matter of the body

$C_e$ = Specific heat of the specific matter of the body

$\Delta T$ = Temperature of the specific matter of the body

$C_e = \frac{248.2}{6\Delta T}$

$C_e = \frac{248.2}{6 \times 116}$

$C_e$ = 0.356 Joule per gram degree centigrade.

An unspecified metal is store into the laboratory.  The weight of the unspecified metal is 6 gram. Now 248.2 Joule energy is added into the unspecified metal. The temperature of the unspecified metal is rise up to 116 degree centigrade. The amount of specific heat for the unspecified metal is 0.356 Joule per gram degree centigrade.

Conclusion:

Two physical systems are in thermal equilibrium if there is no net flow of thermal energy between them when they are connected by a path permeable to heat.

Indrani Banerjee

Hi..I am Indrani Banerjee. I completed my bachelor's degree in mechanical engineering. I am a enthusiastic person and I am a person who is positive about every aspect of life. I like to read Books and listening to music.