A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL of pure water. the temperature of the water changed from 21 degrees C to 23.68 degrees C. what is the specific heat of the metal? specific heat of water is 4.184 J/g C.

0.111 J/g°C

Explanation:

We are given;

Mass of the unknown metal sample as 58.932 g Initial temperature of the metal sample as 101°C Final temperature of metal is 23.68 °C Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

Therefore; mass of pure water is 45.2 g Initial temperature of water = 21°C Final temperature of water is 23.68 °C Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

Step 1: Calculate the amount of heat gained by pure water

Q = m * c * ΔT

For water, ΔT = 23.68 °C - 21° C

= 2.68 °C

Thus;

Q = 45.2 g * 4.184 J/g°C * 2.68°C

= 506.833 Joules

Step 2: Heat released by the unknown metal sample

We know that, Q = m * c * ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

= 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g * c * 77.32°C

= 4556.62c Joules

Step 3: Calculate the specific heat capacity of the unknown metal sample We know that, the heat released by the unknown metal sample is equal to the heat gained by the water. Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 : 4556.62

= 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C