A cup holding 125.12g of water has an initial temperature of 26.8 degrees C. After a 35.08g piece of metal, at 99.5 degrees C, is transferred to the water, the temperature rises to 29.7 degrees C. What is the specific heat of the metal?

0.620 J/g°C

Explanation:

Heat gained or absorbed, Q by a substance is calculated by;

Q = mass * specific heat capacity * Change in temperature

In this case we are given;

Mass of water = 125.12 g Initial temperature of water = 26.8 °C Initial temperature of the metal = 99.5°C Final temperature of the mixture = 29.7 °C

We are required to calculate the specific heat capacity of the metal;

Step 1 : Heat absorbed by water

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

Temperature change of water = 29.7 °C - 26.8°C

= 2.9 °C

But, Q = m*c*ΔT

Thus, Heat = 125.12 g * 2.9°C * 4.184 J/g°C

= 1518.156 Joules

Step 2; Heat lost by the metal

Specific heat capacity of the metal = x J/g°C

Temperature change of the metal = 29.7 °C - 99.5°C

= - 69.8 °C

But, Q = mcΔT

Therefore;

Heat lost by the meatl = 35.08 g * x J/g°C * 69.8 °C

= 2.448.584x Joules

Step 3: C; aculating the specific heat capacity of the metal

The heat gained by water is equal to the heat lost by the metal

Therefore;

1518.156 Joules = 2.448.584x Joules

x = 1518.156 J : 2.448.584 J

= 0.620 J/g°C

Therefore, the specific heat of the metal is 0.620 J/g°C