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.
The hotter star will be 16 times more luminous - luminosity depends on two things - the size of the star and the temperature of the star. The hotter a star is, the more energy it will give out. This will give rise to greater luminosity.
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From A to B its 5 ohm.
above shown 6 and 12 ohm resistors are in parallel to short circuit hence their equivalent resistance is zero.
(Current doesnt flow through a resisstor if there is a Short circuit alternate.
10/9
Explanation:
option 2 is the correct answer
