How many watts would I need to get CO2 from 30 °C to 4000 °C in 1 hour Scenario: After reading some articles on new sorts of ways to deal with $\ce{CO2}$ in the atmosphere, a question came to mind: If I had 100 kg of $\ce{CO2}$, and I would like to heat it up from 30 °C to 4000 °C in a given period (lets say 1 hour), how much energy would I need? What would that be in watts? And from there, assuming a solar panel with 20 % efficiency for 1 m² (therefore 200 watts/hour), what would be the area of solar panels needed to reach heat the $\ce{CO2}$?

Obs: From my very forgotten knowledge of Chemistry from high school, I found the specific heat for $\ce{CO2}$ should be 1000 to 1800 kJ/(kg K), and from this page http://www.heatweb.com/techtips/useful.html I got a formula and tried calculating it, but my units are probably all wrong, because the value was considerably low. I then multiplied 100 kg * 1500000 * 3970 °C. With the energy value, I divided by time (1 hour) to have the output in watts.

Obs2: I know that I am forgetting or do not know about many other limitations (such as the dissociation above a certain temperate). I also know that it would be inefficient and very costly, but would nonetheless like to find out the values, to try and figure what would be necessary for this procedure.

Question: Is this a viable way to do this calculation? What else should I consider?
I approached this in a similar way utilizing the heat capacity of $\ce{CO2}$ over a temperature range.

An estimation can be made by approximating the change in enthalpy of $\ce{CO2(g)}$ over a given temperature range. This can be accomplished through the integral form for changes in Enthalpy:

$$\Delta{H} = \bar{H_1} - \bar{H_2} = \int_{T_1}^{T_2}C_p(T)\,\mathrm dT$$

Which has units of [energy/mass].

Often times, the function $C_p(T)$ for certain gases and liquids will be tabulated in textbooks or chemical databases (such as NIST). In this case, I will utilize a function for the $C_p(T)$ of $\ce{CO2}$ in two temperature ranges (300–1200 K, 1200–6000 K)$^1$:

$$C_p(T)=A+BT+CT^2+DT^3+\frac{E}{T^2}$$

Range 1 (300–1200 K) coefficients:

$A=24.997,$ $B=55.187,$ $C=-33.691,$ $D=7.948,$ $E=-0.137$

Range 2 (1200–6000 K) coefficients:

$A=58.166,$ $B=2.720,$ $C=-0.492,$ $D=0.039,$ $E=-6.447$

Next, we can obtain the energy required for this temperature change by multiplying $\Delta{H}$ by the mass of $\ce{CO2}$ we're interested in, such that:

$$Q = m_\ce{CO2}\,\Delta{H} =  m_\ce{CO2}\int_{T_\text{initial}}^{T_\text{final}}C_p(T)\,\mathrm dT = m_\ce{CO2}\left[\int_{T_1}^{T_2}C_p(T)\,\mathrm dT + \int_{T_2}^{T_3}C_p(T)\,\mathrm dT\right]$$

Evaluating the integrals (keeping in mind that A, B, C, and D will change based on the temperature range in this case) yields:

$$m_\mathrm{CO2}\times7.266\times10^{12}\ \mathrm{\frac{J}{mol}}$$

$$\frac{100}{0.04401}\ \mathrm{\frac{kg}{\frac{kg}{mol}}}\times7.266\times10^{12}\ \mathrm{\frac{J}{mol}} = 1.651\times10^{16}\ \mathrm J$$

In watts, this value becomes:

$$\left(1.651\times10^{16}\ \mathrm J\right) \left(\frac{1}{3600}\ \mathrm{\frac{h}{s}}\right) = 4.560\times10^{12}\ \mathrm W$$ 

Calculating the solar-panned area required:

$$A_\text{panels} = \frac{4.560\times10^{12}}{200}\ \mathrm{\frac{W}{\frac{W}{m^2}}} = 2\times10^{10}\ \mathrm{m^2}$$

Source [1]: “Carbon Dioxide.” Carbon Dioxide, National Institute of Standards and Technology, webbook.nist.gov/cgi/cbook.cgi?ID=C124389&amp;Mask=1#Thermo-Gas.