An object is freely falling from a cliff. If this object has an initial velocity of 3 m/s and completes its fall have a speed of 100 m/s, how high is this cliff?
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Ответ:
parte a) el calor específico es c = 0.383 J /(gr*K)
parte b) la temperatura inicial es T inicial= 52.72 °C
Explanation:
para el primer punto la formula para el calor Q es:
Q = m * c * ( T final - T inicial )
donde
m= masa de la pieza = 0.300 kg = 300 gr
Q = flujo de energía en forma de calor= 2299 J
c = calor específico
T final = temperatura final =40°C
T inicial = temperatura inicial = 60 °C
entonces
Q = m * c * ( T final - T inicial )
c = Q / [ m* ( T final - T inicial ) = 2299 J/[ 300 gr * ( 60 °C - 40°C )]
= 0.383 J /(gr*K)
c = 0.383 J /(gr*K)
para el segundo punto usamos la misma formula
Q = m * c * ( T final - T inicial )
pero
m= 200 gr= 0.200 kg
c=459.8 J/(kg*K) , Q =20.900 J , T final = 280 °C
Q = m * c * ( T final - T inicial )
T inicial = T final - Q/(m*c) =280 °C - 20.900 J/(459.8 J/(kg*K)* 0.200 kg) = 52.72 °C