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### Course: Physical Chemistry (Essentials) - Class 11>Unit 6

Lesson 4: Measurement of U and H- Calorimetry

# Heat capacity

Heat capacity is the amount of heat required to change the temperature of a given amount of matter by 1°C. The heat capacity of 1 gram of a substance is called its specific heat capacity (or specific heat), while the heat capacity of 1 mole of a substance is called its molar heat capacity. The amount of heat gained or lost by a sample (q) can be calculated using the equation q = mcΔT, where m is the mass of the sample, c is the specific heat, and ΔT is the temperature change. Created by Jay.

## Want to join the conversation?

• At Jay says that it doesn't rally matter what units you use here, why is that? If I wanted to convert it from Celsius to kelvin would I still add 273 to the temperature?
• We're dealing with change in temperatures rather than single temperatures. And a change in a degree Celsius is the same as a change in Kelvin; an increase of 1 degree Celsius is the same as an increase of 1 kelvin. Therefore the delta T (and the specific heat capacity) can use units of Celsius or kelvin.

So for the example at , the ΔT could be 11K or 11 degrees C since a change in either is equivalent.

So if they wanted the initial or final temperature (a single temperature value, not a change), you would have to use a conversion depending which units you wanted.

Hope that helps.
• why does jay write 1.000 g instead of 1 g?
• Jay includes those zeros because it reflects greater precision of the instrument he used to measure the mass of the water. In science we want to report as many significant figures as is allowed by our instrumentation, no more no less. So mathematically we could have wrote 1 g and the answer numerically would the same, but we would have sacrificed our precision. In science we have to be conscious of things like the limitations of measurements due to instruments and uncertainty associated with those measurements.

Hope that helps.
• What is q in the equation of heat capacity , it can't be the heat required since the heatrequired is the heat capacity itself
Help
• Q is the amount of supplied or subtracted heat in Joules
• at , why was did he add 75.2J in 1 mol of water to increase temp by 1 degree? is the energy added simply Cm times the number of mols of water?
• Jay is essentially doing what he did at , but with 1 mol of water and the molar heat capacity instead of 1 gram of water and the specific heat capacity.

Hope that helps.
• At -, Jay says that it equals 22K or 22 degrees Celsius. Why is this? Why wouldn't it be -251 degrees Celsius, or something like that? You convert between Kelvin and Celsius by adding or subtracting 273, right? Thanks!
• The size of each unit in the Celsius scale is equivalent to each unit in a Kelvin scale. The only difference is that the Kelvin scale is shifted (or vice versa).

If we were calculate T, a temperature, you would be right to assume that we should take this into account. However, we are calculating ΔT. Notice that (x1 + 273) - (x2 + 273) is in fact equal to (x1) - (x2). In other words, the extra "273" in the Kelvin scale is canceled out and the only bit of importance is the actual difference in temperature.

It is worth noting that this would not work for the Fahrenheit scale since each unit in it is smaller.

Hope this helps.
• Can the value of a substance's heat capacity also mean if the temperature of the object decreases by 1º C or 1 K, that amount of heat is transferred to the surroundings? For example, if 1 gram of water decreases its temperature from 15.5ºC to 14.5ºC, does it mean 4.18 J of energy is released in the surroundings?

Also, I'm still having trouble understanding why the heat capacity can exchange ºC or K for the units of temperature. Why is it that a change in degree Celsius is the same as a change in Kelvin (isn't the change scaled by 273 K for 0ºC)?
• Yeah your first assumption is correct. If you work out the math then the heat associated with a negative change in temperature gives the heat a negative sign. It'll have the same magnitude as the same increase in temperature, just different signs. So yeah it can be thought of as the amount of energy required to raise the temperature, and also the amount of energy lost by a decrease in temperature.

For the second question, we can exchange the temperature part of heat capacity's unit because the equation, Q = (M)(C)(ΔT), uses the change in temperature as opposed to a single temperature value. If we're using a single temperature, then it does matter which temperature scale and unit we're using. However here with a change in temperature it doesn't matter since the Celsius and Kelvin scales are defined such that a change in one scale is equivalent to the other. Mathematically you can imagine this as the two scales being lines having the same slopes, only differing in their y-intercepts. So since a change in Celsius is the same as a change in Kelvin, we can use either unit.

For example, individual temperatures like 25°C or 28°C are different from 298K and 301K, but the ΔT is the same for both scales (°C: 28-25 = 3 and K: 301-298 = 3).

Hope that helps.
• @ in the video he just says "when we do the math" and doesn't really go into or explain the math, i dont understand how from: 1.0*10^2=10.0g(0.90J/g-K) could produce the value 11 for the change in temperature.
• It’s an algebra problem where we’re solving for ΔT in the heat capacity equation.

So, q = mCΔT, given equation
q/(mC) = ΔT, divide by m and C for both sides of the equation. So ΔT is solved for then we substitute the values in to get a numerical answer.

(1.0 x 10^(2))/(10.0*0.90) = 11.111… or 11 when rounded for sig figs.

Hope that helps.
• Shouldn't the change in mass also be taken into account when calculating heat capacity?
• the equation assumes that mass is constant. We're calculating the amount of energy it takes to change the temperature of a given and constant mass.
(1 vote)
• in our book waters specific heat is written as 4200.0jper kg kelvin but here it is 4.18jper gram Celsius can somebody explain why?