Formula for calculating the amount of heat
Let’s say you need to find out how much heat an iron part received when heated. The mass of the part is $3 \space kg$. The part heated from $20 \degree C$ to $300 \degree C$.
Let's take the value of the heat capacity of iron from the table - $460 \frac{J}{kg \cdot \degree C}$. Let us explain the meaning of this value: to heat a piece of iron weighing $1 \space kg$ by $1 \degree C$ it is necessary to expend an amount of heat equal to $460 \space J$.
- The mass of the part is 3 times greater, which means that heating it will require 3 times the amount of heat - $1380 \space J$
- The temperature changed not by $1 \degree C$, but by $280 \degree C$
- This means that 280 times more heat is needed: $1380 \space J \cdot 280 = 386,400 \space J$
Then, the formula for calculating the amount of heat required to heat a body or released by it during cooling will take the form:
$Q = cm(t_2 - t_1)$,
where $Q$ is the amount of heat, $c$ is the specific heat capacity of the substance of which the body is composed, $m$ is the mass of the body, $t_1$ is the initial temperature of the body, $t_2$ is the final temperature of the body.
To calculate the amount of heat that must be expended to heat a body or released by it during cooling, the specific heat capacity must be multiplied by the mass of the body and by the difference between the final and initial temperatures.
Let us consider in more detail the features of calculating the amount of heat using examples of problem solving.
Specific heat capacity of a substance
It is a physical quantity that expresses the amount of heat required by a substance per unit mass to increase its temperature by one unit.
Thus, specific heat capacity is a property of a substance because its value is representative of each substance, each of which in turn has different values depending on what state it is in (liquid, solid, or gas).
Specific heat capacity is denoted by the small letter c and is measured in J/kg∗°C, and is the coefficient of temperature increase in one unit of the entire system or the entire mass of a substance.
In addition, specific heat capacity varies depending on the physical state of a substance, especially in the case of solid particles and gases, since its molecular structure affects heat transfer in the particle system. The same applies to atmospheric pressure conditions: the higher the pressure, the lower the specific heat.
The basic composition of the specific heat of a substance should be c = C/m, i.e., the specific heat is equal to the ratio of calorie content and mass.
However, when this is applied to a given temperature change, it is said to be average specific heat capacity, which is calculated based on the following formula:
Where:
Q is the transfer of thermal energy between the system and the environment (J);
m is the mass of the system (kg);
Δt or (t2 - t1) is the increase in temperature to which it is exposed (°C).
Formula for finding the amount of heat Q:
Q = c∗m(t 2
- t 1 )
The higher the specific heat capacity of a substance, the more thermal energy is required for its temperature to increase. For example, heating water (lead = 4200 J/kg∗°C) will require more thermal energy than heating lead (lead = 140 J/kg∗°C).
Heat balance equation:
Q given + Q received = 0.
Below is a table of specific heat capacity values for some substances:
Calculation of the amount of heat spent on heating two bodies
Water with a mass of $10 \space kg$ was poured into an iron pot with a mass of $4 \space kg$ (Figure 1). Their temperature is $25 \degree C$. How much heat is needed to heat the pot and water to a temperature of $100 \degree C$?
Figure 1. Heating water in a pot.
Please note that two bodies will heat up at once : both the pot and the water in it. There will be constant heat exchange . Therefore, we can consider their temperatures to be the same.
Note that the masses of the pot and water are different. They also have different heat capacities. This means that the amounts of heat they receive will be different .
Now we can write down the condition of the problem and solve it.
Given: $m_1 = 4 \space kg$ $c_1 = 460 \frac{J}{kg \cdot \degree C}$ $m_2 = 10 \space kg$ $c_2 = 4200 \frac{J}{kg \cdot \ degree C}$ $t_1 = 25 \degree C$ $t_2 = 100 \degree C$
Q-?
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Solution:
To calculate the amount of heat received, we use the formula $Q = cm(t_2 - t_1)$.
Let's write this formula for the amount of heat received by the boiler: $Q_1 = c_1m_1(t_2 - t_1)$.
Let's calculate this amount of heat: $Q_1 = 460 \frac{J}{kg \cdot \degree C} \cdot 4 \space kg \cdot (100 \degree C - 25 \degree C) = 1840 \frac{J}{\ degree C} \cdot 75 \degree C = 138,000 \space J = 138 \space kJ$.
The amount of heat received by water when heated will be equal to: $Q_2 = c_2m_2(t_2 - t_1)$.
Let's substitute the numerical values and calculate: $Q_2 = 4200 \frac{J}{kg \cdot \degree C} \cdot 10 \space kg \cdot (100 \degree C - 25 \degree C) = 42000 \frac{J}{ \degree C} \cdot 75 \degree C = 3,150,000 \space J = 3150 \space kJ$.
The total amount of heat spent on heating the boiler and water: $Q = Q_1 +Q_2$, $Q = 138 \space kJ + 3150 \space kJ = 3288 \space kJ$.
Answer: $Q = 3288 \space kJ$.
Examples of problem solving
The following problems will show examples of calculating the required amount of heat.
Task No. 1
How much heat is needed to produce steam at 100°C from ice weighing 2 kg, taken at a temperature of -10°C?
Solution:
Answer: in order to obtain steam at 100°C from ice weighing 2 kg, taken at a temperature of -10°C, you need to take 6.162 megajoules of heat.
Task No. 2
An iron cauldron weighing 5 kg is filled with water weighing 10 kg. How much heat must be transferred to the boiler with water to change its temperature from 10 to 100°C?
Let's start the solution and note that both the boiler and the water will heat up. The temperature difference will be 1000C - 100C = 900C. That is, the boiler temperature will change by 90 degrees, and the water temperature will also change by 90 degrees.
The amounts of heat that both objects received (Q1 for the boiler and Q2 for the water) will not be the same. We will find the total amount of heat using the heat balance formula Q = Q1 + Q2.
Calculation of the amount of heat when mixing liquids
Hot water was diluted with cold water and the temperature of the mixture was $30 \degree C$. Hot water with a temperature of $100 \degree C$ was $0.3 \space kg$. The cold water had a mass of $1.4 \space kg$ and a temperature of $15 \degree C$. Calculate how much heat was given up by hot water when cooling and received by cold water when heated. Compare these amounts of heat.
Given: $c_1 = c_2 = c = 4200 \frac{J}{kg \cdot \degree C}$ $m_1 = 0.3 \space kg$ $m_2 = 1.4 \space kg$ $t_1 = 100 \degree C$ $t_2 = 15 \degree C$ $t = 30 \degree C$
$Q_1 — ?$ $Q_2 — ?$
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Solution:
Let's write a formula for calculating the amount of heat given off by hot water when cooling from $100 \degree C$ to $30 \degree C$: $Q_1 = cm_1(t_1 - t)$.
Let's calculate this value: $Q_1 = 4200 \frac{J}{kg \cdot \degree C} \cdot 0.3 \space kg \cdot (100 \degree C - 30 \degree C) = 1260 \frac{J}{\degree C} \cdot 70 \degree C = 88 200 \space J = 88.2 \space kJ$.
Let's write a formula for calculating the amount of heat received by cold water when heated from $15 \degree C$ to $30 \degree C$: $Q_2 = cm_2(t - t_2)$.
Let's calculate this value: $Q_1 = 4200 \frac{J}{kg \cdot \degree C} \cdot 1.4 \space kg \cdot (30 \degree C - 15 \degree C) = 5880 \frac{J}{\degree C} \cdot 15 \degree C = 88 200 \space J = 88.2 \space kJ$.
$Q_1 = Q_2 = 88.2 \space kJ$.
Answer: $Q_1 = Q_2 = 88.2 \space kJ$.
In the course of solving this problem, we saw that the amount of heat given off by hot water and the amount of heat received by cold water are equal. Other experiments give similar results.
Means,
If heat exchange occurs between bodies, then the internal energy of all heating bodies increases by as much as the internal energy of cooling bodies decreases.
In practice, it often turns out that the energy released by hot water is greater than that received by cold water. In fact, when hot water cools, it transfers some of its internal energy to the air and the vessel in which mixing occurs.
There are 2 ways to take this factor into account:
- If we reduce energy losses as much as possible, we will achieve an approximate equality of energy given and received
- If we calculate and take into account energy losses, we can obtain the exact equality
Types of Heat Transfer
Heat transfer is the process of heat transfer (energy exchange).
Everything here is quite simple, there are only three types: thermal conductivity, convection and radiation.
Thermal conductivity
That type of heat transfer that can be characterized as the ability of bodies to conduct energy from a more heated body to a less heated one.
It's about transferring heat through contact. Admit it, have you ever warmed yourself near a radiator? If you sat close to it, then you warmed up due to thermal conductivity. Cuddling a cat who has a hot belly is also effective.
Sometimes we go a little overboard with the possibilities of this effect when we lie down on the hot sand at the beach. There is an effect, but not a very pleasant one. Well, an ice heating pad on your forehead has the opposite effect - your forehead transfers heat to the heating pad.
Convection
When we talked about thermal conductivity, we used a battery as an example. Conduction is when we get heat by touching a battery. But all the things in the room do not touch the radiator, and the room is heated. This is where convection comes in .
The fact is that cold air is heavier than hot air (cold air is simply denser). When the battery heats a certain volume of air, it immediately rises to the top, passes along the ceiling, has time to cool down and go back down to the battery, where it heats up again. Thus, the entire room is heated evenly, because increasingly hot currents replace increasingly colder ones.
Radiation
We already mentioned the beach, but we were talking only about the hot sand. But the heat from the sun is radiation . In this case, heat is transferred through waves.
If we warm ourselves by the fireplace, do we receive heat by convection or radiation?
