An aluminum calorimeter with a mass of 100 g contains 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 100.0-g piece of copper at 50.0°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 specific heat of the unknown sample.
(b) Using the data in the table above, can you make a positive identification of the unknown material?
(c) Can you identify a possible material?
(d) Explain your answers for part (b).

a)

b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c) The material is peat, possibly.

d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.

Explanation:

Given:

We have,

specific heat of aluminium,

specific heat of copper,

specific heat of water,

Using the heat energy conservation equation.

The heat absorbed by the system of the calorie-meter to reach the final temperature.

The heat released by the blocks when dipped into water:

where

specific heat of the unknown material

For the conservation of energy :

so,

b)

This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c)

The material is peat, possibly.

d)

The material cannot be ice because ice doesn't exists at a temperature of 100°C.

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5