If a 50.00 kg sample of copper (0.385 J/(g·K)) initially at 5.0 oC is struck by 2.50 moles of 364 nm photons, and all of the energy of the..?

Let T°C be the final temperature.

Energy released by the photons
= N h ν
= N h c / λ
= (6.022 × 10²³) × (6.626 × 10⁻³⁴) × (3.00 × 10⁸) / (364 × 10⁻⁹) J

Energy absorbed by the copper
= m c ΔT
= (50.00 × 10³) × 0.385 × (T - 5) J

Energy absorbed by the copper = Energy released by the photons
(50.00 × 10³) × 0.385 × (T - 5) = (6.022 × 10²³) × (6.626 × 10⁻³⁴) × (3.00 × 10⁸) / (364 × 10⁻⁹)
T = 22.1

Final temperature = 22.1°C