A sample of aluminum with a mass of 12.0 grams is heated in boiling water to a t

Question

A sample of aluminum with a mass of 12.0 grams is heated in boiling water to a temperature of 96.0 degree C. The ho metal is transferred to a sample of water with a mass of 20.0 grams and the metal cools to a temperature of 29 0 degree C. Calculate the heat released by the aluminum metal. The heat released by the metal increases the temperature of the water. Use the value of the heat released by the metal to calculate the temperature change of the water. What is the change in temperature of the water? 10.8 degree C 726 degree C 60.5 degree C 8.68 degree C

Explanation / Answer

mal = 12 g

Tal = 96

mw = 20 g

Tfinal = 29

So...

Heat of aluminium = - Heat of water

mal*Cal*(Tf-Tal) = -mw*Cw*(Tf-Tw)

-12*0.91*(29-96) = 20*4.184*(29-Tw)

solve for Tw

731.64 = 20*4.184*(29-Tw)

Tw = -(731.64 /(20*4.184) - 29)

Tw= 20.256

change of water = 29-20.256 = 8.744 °C

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