Question

An aluminum calorimeter with a mass of 100 g con- tains 250 g of water. The calorimeter and water are in thermal equilibrium at 10.0°C. Two metallic blocks are placed into the water. One is a 50.0-g piece of copper at 80.0°C. The other has a mass of 70.0 g and is originally at a temperature of 100°C. The entire system stabilizes at a final temperature of 20.0°C. (a) Determine the spe- cific heat of the unknown sample.

Answer 1

Answer:1.587J / g degC.

Explanation:SOLUTION

Specific heat of aluminum = 0.897 J / g degC.

Heat gained by calorimeter = 100 * 10 * 0.897 J = 897 J

Specific heat of water = 4.18 J / g degC.

Heat gained by water = 250 * 10 * 4.18 J = 10450 J

Specific heat of copper = 0.385 J / g degC.

Heat loss by copper block = 50 * (80-10) * 0.385 J =1347.5 J

Let specific heat of unknown block be x J / g degC.

Heat loss by unknown block = 70 * (100-10) * x J = 6300x J

Heat gain = Heat loss

897 + 10450 = 1347.5 + 6300x

6300x = 9999.5

x =1.58722= 1.587

Specific heat of unknown substance is 1.587J / g degC.