Question

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)

Answer 1

1.114 J/g°C

Step-by-step 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