An aluminum cup of mass 150 g contains 800 g of water in thermal equilibrium at 80.0°C. The combination of cup and water is cooled uniformly so that the temperature decreases by 1.50°C per minute. At what rate is energy being removed by heat? Express your answer in watts.

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