An automobile tire having a temperature of 3.4 ◦C (a cold tire on a cold day) is filled to a gauge pressure of 24 lb/in2 . What
2 answers:
Answer:
27.16 lb/in²
Explanation:
initial temperature, T1 = 3.4 °C = 276.4 K
initial gauge pressure, P1 = 24 lb/in²
atmospheric pressure, Po = 14.7 lb/in²
initial absolute pressure, P1' = Po + P1 = 14.7 + 24 = 38.7 lb/in²
final temperature, T2 = 26 °C = 299 K
Let the final gauge pressure is P2.
use the ideal gas equation and the volume is constant.
P2' = 41.86 lb/in²
Now the gauge pressure, P2 = P2' - Po = 41.86 - 14.7 = 27.16 lb/in²
Thus, the new gauge pressure is 27.16 lb/in².
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Answer:
The final temperature of both objects is 400 K
Explanation:
The quantity of heat transferred per unit mass is given by;
Q = cΔT
where;
c is the specific heat capacity
ΔT is the change in temperature
The heat transferred by the object A per unit mass is given by;
Q(A) = caΔT
where;
ca is the specific heat capacity of object A
The heat transferred by the object B per unit mass is given by;
Q(B) = cbΔT
where;
cb is the specific heat capacity of object B
The heat lost by object B is equal to heat gained by object A
Q(A) = -Q(B)
But heat capacity of object B is twice that of object A
The final temperature of the two objects is given by
But heat capacity of object B is twice that of object A
Therefore, the final temperature of both objects is 400 K.