Question

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.​

Answer 1

0.111 J/g°C

Step-by-step 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