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P1 E) Specific Heat Capacity
P1 E) Specific Heat Capacity
Different substances require different amounts of energy to be transferred to their thermal energy stores in order for their temperature to increase. For example, we need to transfer 4,181 joules of energy to heat 1 kg of water up by 1°C, and we only need to transfer 139 joules of energy to increase 1 kg of mercury by 1°C.
Materials that gain a lot of energy in their thermal energy stores when they heat up will give out more energy when they cool down; 1 kg of water will give out 4,181 joules of energy when it cools down by 1°C, and 1 kg of mercury will only give out 139 joules of energy when it cools down by 1°C.
We can work out the change in energy in a substance’s thermal energy stores by using the formula below:
Materials that gain a lot of energy in their thermal energy stores when they heat up will give out more energy when they cool down; 1 kg of water will give out 4,181 joules of energy when it cools down by 1°C, and 1 kg of mercury will only give out 139 joules of energy when it cools down by 1°C.
We can work out the change in energy in a substance’s thermal energy stores by using the formula below:
In the above formula, ΔE is the change in thermal energy in joules (J), m is the mass of the substance in kg, c is the specific heat capacity of a particular substance in J/kg°C and Δθ is the change in temperature in °C.
c in the above formula is the specific heat capacity, which is the amount of energy required to increase the temperature of 1 kilogram of a substance by 1°C. Like I said earlier, different substances require different amounts of energy to be transferred to their thermal energy stores in order to increase their temperature. The specific heat capacity is measure is J/kg°C. The specific heat capacity of water is 4,181 J/kg°C, the specific heat capacity of oxygen is 918 J/kg°C and the specific heat capacity of lead is 128 j/kg°C. You do not need to remember the specific heat capacities of any substances for the exams – you will always be given them.
A substance that’s temperature increases will have more energy in its thermal energy store. A substance that’s temperature decreases will have less energy in its thermal energy store. We can use this specific heat capacity formula to calculate the increase or decrease in energy in the thermal energy stores of a substance.
c in the above formula is the specific heat capacity, which is the amount of energy required to increase the temperature of 1 kilogram of a substance by 1°C. Like I said earlier, different substances require different amounts of energy to be transferred to their thermal energy stores in order to increase their temperature. The specific heat capacity is measure is J/kg°C. The specific heat capacity of water is 4,181 J/kg°C, the specific heat capacity of oxygen is 918 J/kg°C and the specific heat capacity of lead is 128 j/kg°C. You do not need to remember the specific heat capacities of any substances for the exams – you will always be given them.
A substance that’s temperature increases will have more energy in its thermal energy store. A substance that’s temperature decreases will have less energy in its thermal energy store. We can use this specific heat capacity formula to calculate the increase or decrease in energy in the thermal energy stores of a substance.
Example 1
Work out the change in energy in the thermal energy store of 4 kg of water if we increase the temperature of the 4 kg of water by 15°C.
The specific heat capacity of water is 4,181 J/kg°C.
We can work out the change in thermal energy by using the formula below:
Work out the change in energy in the thermal energy store of 4 kg of water if we increase the temperature of the 4 kg of water by 15°C.
The specific heat capacity of water is 4,181 J/kg°C.
We can work out the change in thermal energy by using the formula below:
We are looking for the change in thermal energy, which is ΔE in the formula above. We can find the change in thermal energy (ΔE) by subbing the values for mass (m), specific heat capacity (c) and change in temperature (Δθ) into the formula. We are told in the question that the mass of water is 4 kg, the specific heat capacity of water is 4,181 J/kg°C and the change in temperature is 15°C. All of these values are in the correct units, so we sub them into the formula.
Therefore, the change in energy in the thermal energy store of the 4 kg of water is 250,860 joules.
Example 2
Calculate the change in energy in the thermal energy store of 600 grams of magnesium if we increase its temperature from 40°C to 90°C. Give your answer in kilojoules (kJ)
The specific heat capacity of magnesium is 1,024 J/kg°C.
Like the previous question, we find the change in energy in the thermal energy store of the 600 grams of magnesium by using the formula below:
Calculate the change in energy in the thermal energy store of 600 grams of magnesium if we increase its temperature from 40°C to 90°C. Give your answer in kilojoules (kJ)
The specific heat capacity of magnesium is 1,024 J/kg°C.
Like the previous question, we find the change in energy in the thermal energy store of the 600 grams of magnesium by using the formula below:
We are looking for ΔE, which we find by subbing m, c and Δθ into the formula. We need to make sure that all of these are subbed in with the correct units.
We now have everything that we need in the correct units; we sub m in as 0.6 kg, c in as 1,024 J/kg°C and Δθ as 50°C.
- m is the mass measured in kilograms (kg). The question tells us that the mass is 600 grams, which is not in the correct units. There are 1,000 grams in 1 kilogram, so we convert grams to kilograms by dividing by 1,000; therefore, the mass is 0.6 kg (600 ÷ 1,000 = 0.6).
- c is the specific heat capacity of magnesium, which is 1,024 J/kg°C.
- Δθ is the change in temperature. The question tells us that the initial temperature of the magnesium is 40°C and the final temperature of the magnesium is 90°C. This means that the temperature change of the magnesium is 50°C (90 – 40 = 50).
We now have everything that we need in the correct units; we sub m in as 0.6 kg, c in as 1,024 J/kg°C and Δθ as 50°C.
The change in energy in the thermal energy store of the 600 grams of magnesium is 30,720 joules. The question asks us to give the change in energy in kilojoules and not joules. 1 kilojoule is 1,000 joules, so we convert joules to kilojoules by dividing by 1,000.
Therefore, the change in energy in the thermal energy store of the 600 grams of magnesium is 30.72 kj.
Example 3
The two examples that we have looked at have been fairly straightforward because we have just subbed the values into their correct places in the specific heat capacity formula. We are now going to have a look at an example whereby we are asked to find the temperature change of a particular substance. The easiest way to answer a question like this is to sub the values into their appropriate places in the formula, and then solve to find the value of the unknown that we are looking for. We find the value that we are looking for in the same way that we would solve an algebraic maths equation. Let’s have an example.
Question
800 grams of vanadium is at an initial temperature of 50°C. Find the final temperature of the 800 grams of vanadium after we transfer 18,032 joules of energy to its thermal energy store.
The specific heat capacity of vanadium is 490 J/kg°C.
The first step in finding the final temperature of the 800 grams of vanadium is to find the change in temperature (Δθ). We do this by using the specific heat capacity formula, which is shown below.
The two examples that we have looked at have been fairly straightforward because we have just subbed the values into their correct places in the specific heat capacity formula. We are now going to have a look at an example whereby we are asked to find the temperature change of a particular substance. The easiest way to answer a question like this is to sub the values into their appropriate places in the formula, and then solve to find the value of the unknown that we are looking for. We find the value that we are looking for in the same way that we would solve an algebraic maths equation. Let’s have an example.
Question
800 grams of vanadium is at an initial temperature of 50°C. Find the final temperature of the 800 grams of vanadium after we transfer 18,032 joules of energy to its thermal energy store.
The specific heat capacity of vanadium is 490 J/kg°C.
The first step in finding the final temperature of the 800 grams of vanadium is to find the change in temperature (Δθ). We do this by using the specific heat capacity formula, which is shown below.
When we use this formula, we need to make sure that all of the values are in their correct units:
We now sub these values into the formula.
- ΔE is change in thermal energy, which is measured in joules. The question tells us that ΔE is 18,032 J.
- m is the mass, which is measured in kilograms. The question tells us that the mass is 800 grams, so we need to convert this to kilograms, which we do by dividing by 1,000; the mass is 0.8 kg (800 ÷ 1,000 = 0.8).
- c is the specific heat capacity of vanadium, which is 490 J/kg°C.
We now sub these values into the formula.
We find the value of Δθ by solving the equation above as if it was an algebraic equation. The first thing that I am going to do is multiply the right side of the equation, which results in us obtaining 392Δθ on the right.
We want to find the value of Δθ and not 392Δθ. Therefore, we divide both sides of the equation by 392.
This tells us that the change in temperature is 46°C.
The question asked us to find the final temperature of the vanadium. We were told in the question that the initial temperature was 50°C, and we have just found that the temperature change is 46°C. We can find the final temperature by adding the temperature change (46°C) onto the initial temperature (50°C).
The question asked us to find the final temperature of the vanadium. We were told in the question that the initial temperature was 50°C, and we have just found that the temperature change is 46°C. We can find the final temperature by adding the temperature change (46°C) onto the initial temperature (50°C).
This tells us that the final temperature of the 800 grams of vanadium is 96°C.