I need to get my stuff back in my room I don’t have a big deal
Answer:
E = 15×10⁻²⁹ J
Explanation:
Given data:
Frequency of photon = 2.2× 10⁷ Hz
Energy of photon = ?
Solution:
Formula:
E = h.f
h = 6.63×10⁻³⁴ Js
by putting values,
E = 6.63×10⁻³⁴ Js × 2.2× 10⁷ s⁻¹
E = 14.586 ×10⁻²⁹ J
E = 15×10⁻²⁹ J
The energy of photon is 15×10⁻²⁹ J.
Answer:
17 kJ
Explanation:
Calculation for the Calculate the energy required to heat 0.60kg of ethanol from 2.2°C to 13.7°C.
Using this formula
q = mC∆T
Where,
q represent Energy
m represent Mass of substance=0.60kg=600g
C represent Specific heat capacity=2.44J·g−1K−1.
∆T represent change in Temperature=2.2°C to 13.7°C.
Let plug in the formula
q=(0.60 kg x 1000 g/kg)(2.44 J/gº)(13.7°C-2.2°C)
q = (600g)(2.44 J/gº)(11.5º)
q=16.836 kJ
q= 17 kJ (Approximately)
Therefore the energy required to heat 0.60kg of ethanol from 2.2°C to 13.7°C will be 17 kJ
