Question

1. A chemistry student added 225 grams of aluminum at 85.00C to 115 grams of water at 23.00C in a perfect calorimeter. The final temperature of the aluminum-water mixture was 41.40C. Use the student’s data to calculate the specific heat of aluminum in joules/gram0C.

Answer 1

Specific heat of water = 4.18 Joules / gram ⁰C

Heat gained by water = mass of water * specific heat of water *( Final temperature - initial temperature)= 115 grams * 41.8 oules / gram ⁰C * (41.40 23.00) ⁰C = 8844.88 joules

Let us assume specific heat of aluminium is x joules / gram ⁰C

Heat lost by aluminium = mass of aluminium * specific heat of aluminium *( Initial temperature - final temperature)= 225 grams * x joules / gram ⁰C * (85.00 -41.40) ⁰C = 9810* x joules

At thermal equilibrium, heat gained by water = heat lost by aluminium

So, 8844.88 = 9810 * x

Or, x = (8844.88/9810) = 0.90

So, specific heat of aluminium is 0.90 Joules / gram ⁰C