Answer: New temperature T2 = 9.66°C
Explanation:
Given;
Energy added ∆E = 20J
Initial temperature T1 = 0°C
Number of atom = 1.0×10^23
Using the energy equation.
∆E = (3/2)nR∆T
∆T = (2/3)∆E/nR
Where ;
R = universal gas constant = 8.314J/mol-K
n = number of moles = N/Nt
Nt = number of atoms in one mole of gas = 6.022×10^23
n = 1.0 ×10^23/6.022×10^23= 0.166
∆T = (2/3)×20/(0.166×8.314)
∆T = 9.66
T2 = T1+∆T = 0 + 9.66
T2 = 9.66°C
