According to the second law of thermodynamics,
the answer is
<span>4. The entropy of the universe is increasing. </span>
How can I help? What’s the question?
Respuesta:
0,0560 cal / gºC.
Explicación:
Cantidad de calor; (Q)
Q = mcΔt; Δt = t2 - t1
m = masa, c = capacidad calorífica específica; Δt = cambio de temperatura
c de agua = 1 cal / gºC
c de aluminio = 0,22 cal / gºC
QTotal = Q de agua + Q de aluminio
Q de agua = 450 * 1 * (26 - 23) = 1350 cal
Q de aluminio = 60 * 0.22 * (26 - 23) = 39.6 cal
QTotal = 1350 + 39,6 = 1389,6 cal
Calor perdido = calor ganado
QTotal = calor perdido
- 1389,6 = 335,2 * c * (26 - 100)
-1389,6 = −24804,8 * c
c = 1389,6 / 24804,8
c = 0,056021 cal / gºC.
Capacidad calorífica específica de la plata = 0,0560 cal / gºC.
Answer:
Explanation:
Part A) Using
light intensity I= P/A
A= Area= π (Radius)^2= π((0.67*10^-6m)/(2))^2= 1.12*10^-13 m^2
Radius= Diameter/2
P= power= 10*10^-3=0.01 W
light intensity I= 0.01/(1.12*10^-13)= 9*10^10 W/m^2
Part B) Using
I=c*ε*E^2/2
rearrange to solve for E= 
((I*2)/(c*ε))
c is the speed of light which is 3*10^8 m/s^2
ε=permittivity of free space or dielectric constant= 8.85* 10^-12 F⋅m−1
I= the already solved light intensity= 8.85*10^10 W/m^2
amplitude of the electric field E= (9*10^10 W/m^2)*(2) / (3*10^8 m/s^2)*(8.85* 10^-12 F⋅m−1)
---> E= (1.8*10^11) / (2.66*10^-3) = (6.8*10^13) = 8.25*10^6 V/m
