Answer:
T ’’ = 293º C
Step-by-step explanation:
Here we must equal the kinetic energy with the thermal energy of the Boltzmann equation
E = K
E = k T
Where k is the Boltzmann constant and T the absolute temperature
K = k T
T ’= 10ºC, the equation to move to Kelvin degrees is
T = (T ’+273) [K]
Let's use the first data
K10 = k T1
We write the second data
2 (K10) = k T2
Let's divide the two equations
2 (K10) / (K10) = k T2 / k T1
2 = T2 / T1
T2 = 2 T1
Now let's write this value in degrees Celsius
(T ’’ + 273) = 2 (T ’+ 273)
T ’’ = 2T ’+ 2 273 -273
T ’’ = 2 T ’+ 273
T ’’ = 2 10 + 273
T ’’ = 293º C
This is the temperature to double the kinetic energy