suppose that you had an ohmmeter and a mystery component that could be either an inductor or a capacitor. How could you use the
1 answer:
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Answer:
Heat Flow Rate : ( About ) 87 W
Explanation:
The heat flowing out of the system each minute, will be represented by the following equation,
Q( cup ) + Q( water ) = m( cup ) c( al )
ΔT + m( w )
c( w )
ΔT
So as you can see, the mass of the aluminum cup is 150 grams. For convenience, let us convert that into kilograms,
150 grams = .15 kilograms - respectively let us convert the mass of water to kilograms,
800 grams = .8 kilograms
Now remember that the specific heat of aluminum is 900 J / kg K, and the specific heat of water = 4186 J / kg
K. Therefore let us solve for " the heat flowing out of the system per minute, "
Q( cup ) + Q( water ) = .15 ( 900 J / kg
K )
1.5 + .8
( 4186 J / kg
K )
1.5,
Q( cup ) + Q( water ) = 5225.7 Joules
And the heat flow rate should be Joules per minute,
5225.7 Joules / 60 seconds = ( About ) 87 W