# Heat transfer calculated from the specific heat formula

Say I have 10g of silver, whose specific heat is known to be 0.235. I've heated it up from 50.0C to 60.0C. How much heat has been transferred?

The equation is:

$$Q = C_pm\Delta t$$

where Cp is the specific heat value, m is the mass of the object, and Δt is the change in temperature in Kelvin.

I got the answer 23.5j, which I derived from

$$Q = (0.235)(10)(60.0-50.0) = 23.5j$$

My teacher said that we have to get 10.0 as our delta T then add 273 to make the equation work with Kelvin, which translates to

$$Q = (0.235)(10)((60.0-50.0)+273) = 665.05j$$

Now I don't see her reasoning in this answer since the difference in Kelvin is the same as the difference in Celsius. It also doesn't help that the book had the 665.05j answer. This was a test too, so my wrong answer bugs me.

If I did the equation in Kelvin, I would have arrived at the same answer. My argument was that translating in Kelvin beforehand should work. Apparently, the answer for delta T needs to be scaled.

So, can someone provide a good argument as to which one of these is the correct way to calculate specific heat? Thanks. (I already know what the right answer is and my friends do too, so please give an argument.)

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dmckee already said this but I figure it's worth repeating because we're really really sure. $$60.0^\circ\mathrm{C} - 50.0^\circ\mathrm{C} = 10\text{ K}$$ You're exactly correct that you should get the same answer by converting to Kelvins before subtracting: $$60.0^\circ\mathrm{C} - 50.0^\circ\mathrm{C} = 333.2\text{ K} - 323.2\text{ K} = 10\text{ K}$$ So you do not add 273K to this result; your teacher and the book are wrong.

Degrees Celsius (and Fahrenheit) are funny things, actually. They are only useful for subtraction. The reason is that these temperature systems are defined relative to a fixed point, the triple point of water, at which the temperature is defined to be $T_3 = 0.01^{\circ}\mathrm{C}$. So when you say something is at a temperature of $60.0^\circ\mathrm{C}$, you're really saying that $t - T_3 = 59.9^{\circ}\mathrm{C}$. This means that every temperature expressed in degrees Celsius implicitly depends on the triple point of water.

Obviously, not everything in nature depends on the triple point of water. So we would like to have some way of eliminating that dependence before using temperatures in calculations. You can do this by taking a difference between two temperatures. Suppose you had two temperatures, $t_i$ and $t_f$ (for example, $t_i - T_3 = 49.9^\circ\mathrm{C}$ and $t_f - T_3 = 59.9^\circ\mathrm{C}$). $$t_f - t_i = (t_f - T_3) - (t_i - T_3) = 59.9^\circ\mathrm{C} - 49.9^\circ\mathrm{C} = 10\;\Delta^{\circ}\mathrm{C}$$ Here I've "invented" the unit $\Delta^{\circ}\mathrm{C}$ for a temperature difference, because temperature differences and "relative" temperatures don't work the same way. Notice that a temperature difference doesn't depend on $T_3$ at all. In fact, if we used an entirely different reference value in place of $T_3$, the difference would still be the same.

Once you have a temperature difference, you can multiply it or divide it by other things. You can also add or subtract other temperature differences. This is very similar to things like potential energy, where only the difference between two energies is meaningful, not the actual amounts of energy.

Now, it turns out that there are several important formulas in thermodynamics that involve differences between the actual temperature and a particular reference temperature $T_0$; for example, the thermal energy of noninteracting particles, $$\overline{E} = \frac{3}{2}k_B (T - T_0) = \frac{1}{N}\sum_{i=1}^N\frac{1}{2}m_iv_i^2$$ Based on experiments, you can calculate that $$T_0 = -273.15^\circ\mathrm{C}$$ So evidently, nature assigns some special significance to temperature differences relative to $T_0$: the difference $t - T_0$ is important in some way that no other temperature difference (such as $t - T_3$) is. Based on this result, physicists thought it would make sense to develop a temperature scale which set $T_0 = 0$, so that we wouldn't have to keep subtracting it all the time. The first person to reach this conclusion was Lord Kelvin, thus the thermodynamic temperature scale and its unit were named after him. This is the origin of the Kelvin.

So to summarize, when you have a temperature (not a temperature difference) in degrees Celsius, what you really have is $t - T_3$, and when you have a temperature in Kelvin, what you really have is $t - T_0$. In order to convert a temperature from Celsius to Kelvin, you do this: $$\underbrace{t - T_0}_{\text{in K}} = \underbrace{t - T_3}_{\text{in }^\circ\text{C}} + \underbrace{T_3 - T_0}_{273.15\Delta^\circ\mathrm{C}}$$ i.e. you add 273.15 to the numeric value.

On the other hand, when you have a temperature difference, what you really have is $t_f - t_i$, which doesn't depend on any reference point. So to convert from Celsius to Kelvin, you don't need to do anything.

## Application

Here's how this applies to your example. You have a formula $$Q = C_p m\Delta t = C_p m (t_f - t_i)$$ But you can't plug in for $t_f$ and $t_i$ directly. The only information you have is relative to $T_3$: $$t_i - T_3 = 49.9^\circ\mathrm{C}$$ $$t_f - T_3 = 59.9^\circ\mathrm{C}$$ so you have to stick a couple extra terms into that formula: $$Q = C_p m \bigl[(t_f - T_3) - (t_i - T_3)\bigr]$$ Now you can substitute in your numerical values, $$Q = C_p m \bigl[59.9^\circ\mathrm{C} - 49.9^\circ\mathrm{C}\bigr] = C_p m (10\Delta^\circ\mathrm{C})$$ There's no need to add or subtract anything else.

Alternatively, you could convert the temperatures to Kelvins before plugging them in. Converting to Kelvins means that you now have $$t_i - T_0 = 323.2\text{ K}$$ $$t_f - T_0 = 333.2\text{ K}$$ Again, you have to stick a couple extra terms into the formula: $$Q = C_p m \bigl[(t_f - T_0) - (t_i - T_0)\bigr] = C_p m \bigl[333.2\text{ K} - 323.2\text{ K}\bigr] = C_p m (10\Delta\mathrm{K})$$ By definition, the Kelvin and Celsius scales have degrees of the same size, so $\Delta^\circ\mathrm{C} = \Delta\mathrm{K}$, so these two results are the same. But because of the special properties of the temperature $T_0$, you can also show that $\Delta\mathrm{K} = 1\text{ K}$; in other words, when you're dealing with Kelvins, it's safe to leave off the deltas and not worry too much about when $t$ is a temperature and when it's a temperature difference. That only works for Kelvins, though, not degrees Celsius.

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Put two chalk marks on a meter stick and ask her how far about they are?

If she doesn't get it, ask if the answer depends on how far along the stick you made them or only on the separation.

As a matter of psychology this should almost certainly be done in private, and by someone who has not challenged her on the issue in public. I appreciate that it may be too late for that advice, but I thought I should pass it on because I was kinda stupid that way in my youth (and am only partially cured). ::sigh:: All together class "Every day in every way..."

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The teacher is correct. Look at it this way, what is the bone of contention? you say that since it is the temperature difference it should be the same whatever units we aRE using 1. In the formula for specific heat, the temperature has to be in Kelvin. 2. K= C+273 and not temperature kelvin degree Celsius degree Fahrenheit symbol K °C °F boiling point of water 373.15 100. 212. melting point of ice 273.15 0. 32. absolute zero 0. -273.15 -459.67

Now what is the triple point of water. It is 0.01K The term absolute zero is no ordinary no. NOW the above is an extension on the Carnot;s cycle for heat pumps What we do is instead of taking the high and subtracting the low from it, wetake the diff as delT which gives us Q=(0.235)(10)((60.0−50.0)+273)=665.05j

Let me replasce the del T by 60-50. Now you gonna ask me why not at the beginning, why later. The answer is thermodynamics is based on the Kelvin and not the celsius or Farenheight. If you do it later, you make a mistake.

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Technically, the temperature difference should be in Kelvin but because it is a difference, it can be in Celsius, too. Using the transformation $t[^\circ C]+273=T[K]$, we have $\Delta T = T_2-T_1 = (t_2+273)-(t_1+273) = t_2-t_1 = \Delta t$. – Ondřej Černotík Jan 3 '13 at 21:24

Your teacher is correct... if you convert directly the celsius into Kelvin before you subtract the change in temperature the answer for all the question in terms of change in temp.. i the same ... for example.... 25 deg celsius to 30 deg celsius... it will only give you a 5K if converting immediately.... the correct process is (30 - 25) deg celsius + 273 then that is the change in tenp. in term of kelvin scale

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-1 for misinformation. – centralcharge Jul 2 '13 at 9:03

## protected by Qmechanic♦Oct 2 '13 at 6:50

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