Question

A 28.4 g sample of an unknown metal is heated to 39.4 °C, then is placed in a calorimeter containing 50.0 g of water. Temperature of water increases from 21.00 °C to 23.00 °C. What is the specific heat of this metal in this problem?

Answer 1

Answer:

THE SPECIFIC HEAT OF THE METAL IS 0.8983 J/g °C

Explanation:

In solving the problem, we have to understand that:

Heat lost by the metal = Heat gained by the water in the bomb calorimeter

First is to calculate the heat evolved from the reaction

Heat = mass * specific heat * change in temperature

Mass of water = 50 g

specific heat of water = 4.184 J/g °C

Change in temperature = 23 - 21 = 2 °C

So therefore,

Heat = 50 * 4.184 * 2

Heat = 418.4 J

Next is to solve for the specific heat of the metal;

Heat lost by the metal is the same as the heat gained by water

Heat = mass * specific heat of  metal * change in temperature

Change in temperature = 39.4 °C - 23 °C = 16.4 °C

418.4 = 28.4 * C * 16.4

C = 418.4 / 28.4 * 16.4

C = 418.4 / 465.76

C = 0.8983 J/ g °C

The specific heat of the metal is hence 0.8983 J/g °C