A 25.0-g sample of a metal at 98.0 °C is placed in 50.0 g of

water at 18.0 °C. If the final temperature of the metal and

water is 27.4 °C, what is the specific heat, in J/g °C, of the

metal? (3.5)

1.114 J/g°C

Explanation: In a system components with high temperature lose heat that is equivalent to the heat gained by components with low temperature. In this case; the metal sample will lose heat while the water at lower temperature will gain heat. To calculate the specific heat capacity of the metal, we are going to use the following steps. Step 1: Calculate the heat lost by the meat sample

Mass of the metal sample = 25.0 g

Temperature change from 98°C to 27.4° C = - 70.6°C

Assuming the specific heat capacity of the metal is x, then

Heat lost = Mass * specific heat capacity * change in temp

= 25.0 g * x * - 70.6 °C

= - 1765x joules (-ve indicates that heat was lost)

= 1765x joules

Step 2: Calculate heat gained by water

Mass of water = 50.0 g

Temperature change, from 18.0°C to 27.4°C = 9.4°C

Specific heat capacity of water = 4.184 J/g°C

Therefore,

Heat gained = 50.0 g * 4.184 J/g°C * 9.4°C

= 1966.48 Joules

Step 3: Calculate the specific heat of the metal

We know that heat lost is equal to heat gained.

Therefore;

Heat lost by the metal sample = heat gained by water

1765x J = 1966.48 J

x = 1.114 J/g°C

Therefore, the specific heat capacity of the metal is 1.114 J/g°C