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.
About Kelvins
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.
I don't think that you can assume that $I_c=\ln (p_0)$
A lot of equations have logs of dimensioned quantities in them. It's usually not hard to get rid of the log:
$$
\mathrm{ln}\left(\frac{p_1}{p_2}\right) = I_{c,1}-I_{c,2} - \left(\frac{\Delta H^{\circ}_{v,1}}{RT_1}-\frac{\Delta H^{\circ}_{v,2}}{RT_2}\right) + \mathrm{ln}\left(\frac{T_1^{\frac{\left(C^{g}_{p,1} - C_{p,1}^{l}\right)}{R}}}{T_2^{\frac{\left(C^{g}_{p,2} - C_{p,2}^{l}\right)}{R}}}\right)
$$
It may also be that $I_c$ is just a gas-specific constant, of the form $$I_c=\ln p_0-\frac{\Delta H^{\circ}_{v,0}}{R} -\frac{\left(C^{g}_{p,0} - C_{p}^{l,0}\right)}{RT_0}\mathrm{ln}\left(T_0\right) $$
for some $p_0,T_0$.
You can do the same for the other equations as well. When there's a log of a dimensioned quantity, one of two things is possible:
- The equation has the units implicitly assumed (this is generally not done in physics, but you see it a lot in chemistry)
- The equation is a change equation, and you need to subtract one copy of the equation at a different point from it.
The latter form arises when one takes an indefinite integral in the last step while solving a differential equation (Which is why subtracting is the right way to fix this, since that corresponds to tacking limits onto the integral)
Best Answer
You do not need to use the kelvins for the ideal gas formula. The important thing to know is that $PV \propto nT$ where $T$ is the absolute temperature and can be expressed in any arbitrary linear scale (as long as the temperature is still absolute), including the kelvin and the rankine scale.
However, if you choose the constant of proportionality to be $R$ given in J/(K mol), then the temperature's units must be in kelvins for the equation to make sense. But you're not forced to use such a constant. You could pick up any other constant and you could adjust the temperature scale you're using. In particular you could have picked the constant that would make the temperature in rankine units.
Edit: Since the title of the question refers to "formulas", let me mention a case I've seen up to several papers. With regard to thermoelectric materials, there is a $ZT$ factor that's used to gauge the quality of the material to be used as a thermoelectric one, for example to make a thermoelectric generator. The important thing is that $T$ must be the absolute temperature. Unfortunately many papers mention that $T$ is given in kelvins, while it is not necessarily the case. Fortunately enough, I would say most (good) papers refer to $T$ as being the absolute temperature and do not mention the units.