Asample 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

We are given;

But, density of pure water = 1 g/mL

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

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

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

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

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