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seropon [69]
1 year ago
5

A thermally isolated system is made up of a hot piece of aluminum and a cold piece of copper; the aluminum and the copper are in

thermal contact. The specific heat of aluminum is more than double that of copper. Which object experiences the greater temperature change during the time the system takes to reach thermal equilibrium?
a) It is impossible to tell without knowing the volumes.
b) It is impossible to tell without knowing the masses.
c) The copper experiences a greater temperature change.
d) Neither; both objects experience the same magnitude temperature change.
e) The aluminum experiences a greater temperature change.
Physics
1 answer:
blsea [12.9K]1 year ago
4 0

Answer:

b) It is impossible to tell without knowing the masses.

Explanation:

The temperature change of a substance when it receives/gives off a certain amount of heat Q is given by

\Delta T= \frac{Q}{m C_s}

where

Q is the amount of heat

m is the mass of the substance

Cs is the specific heat capacity of the substance

In this case, we have a hot piece of aluminum in contact with a cold piece of copper: the amount of heat given off by the aluminum is equal to the amount of heat absorbed by the copper, so Q is the same for the two substances. However, we see that the temperature change of the two substances depends on two other factors: the mass, m, and the specific heat, Cs. So, since we know only the specific heat of the two substances, but not their mass, we can't tell which object will experience the greater temperature change.

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2 years ago
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Answer:

a. v1 = 5.06 m/s,  v2 = 3.96 m/s ,  R = 1.27

b. t = 1 hr, 11 min, 26 sec  

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p₀ + ρgh₀ + ½ρv₀² = p₁ + ρgh₁ + ½ρv₁²

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Because it is open the tank so P=0 pa so:

0 Pa + 1000kg/m³ * 9.8m/s² * 0.800m + 0 = 0 + 0 + ½ * 1000kg/m³ * v²

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The ratio on the air is solve using both velocities so:

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dh/dt = -u

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dh/dt = -(1/1e4)*√[(p+9800h)/500]

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