Example 6.3: A 32.5-g cube of aluminum initially at 45.8 °C is submerged into 10

Q: Example 6.3: A 32.5-g cube of aluminum initially at 45.8 °C is submerged into 10

6.3) heat lost by Al = heat gained by water mass of Al*s*DT = mass of water*s*DT 32.5*0.9*(45.8-x) = 105.3*4.184*(x-15.4) x = final temperature = 17.3 C practice : heat lost by metal = heat gained by water mass of Metal*s*DT = mass of water*s*DT 350*x*(100-35.2) = 160*4.184*(35.2-22.4) x = specific heat of metal = 0.38 j/g.c

ID: 556915 • Letter: E

Question

Example 6.3: A 32.5-g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C, what is the final temperature of both substances at thermal equilibrium? ( Assume that the aluminum and the water are thermally isolated from everything else Practice - A hot piece of metal weighing 350.0 g is heated to 100.0 °C. It is then placed into a coffee cup calorimeter containing 160.0 g of water at 22.4 °C. The water warms and the copper cools until the final temperature is 35.2 °C. Calculate the specific heat of the metal (C metal, J/g·°C) and identify the metal. (C or water 4.18 J/g. °c

Explanation / Answer

6.3)

heat lost by Al = heat gained by water

mass of Al*s*DT = mass of water*s*DT

32.5*0.9*(45.8-x) = 105.3*4.184*(x-15.4)

x = final temperature = 17.3 C

practice :

heat lost by metal = heat gained by water

mass of Metal*s*DT = mass of water*s*DT

350*x*(100-35.2) = 160*4.184*(35.2-22.4)

x = specific heat of metal = 0.38 j/g.c

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Example 6.3: A 32.5-g cube of aluminum initially at 45.8 °C is submerged into 10

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