Question

A 12.0 g sample of a metal is heated to 90.0 ◦C. It is then dropped into 25.0 g of water. The temperature of the water rises from 22.5 to 25.0 ◦C. The specific heat of water is 4.18 Jg−1◦C −1 . Calculated the specific heat of the metal. Express your answer in Jg−1◦C −1 .

Answer 1

Answer:

The specific heat of the metal is 0.335 J/g°C

Explanation:

Step 1: Data given

Mass of the metal = 12.0 grams

Initial temperature of the metal = 90.0 °C

Mass of the water = 25.0 grams

Initial temperature of water = 22.5 °C

Final temperature of water (and metal) = 25.0 °C

The specific heat of water = 4.18 J/g°C

Step 2: Calculate the specific heat of the metal

Qgained  = -Qlost

Qwater = -Qmetal

Q= m*c*ΔT

m(metal) *c(metal)*ΔT(metal) = -m(water)*c(water)*ΔT(water)

⇒ mass of the metal = 12.0 grams

⇒ c(metal) = TO BE DETERMINED

⇒ ΔT(metal) = T2 - T1 = 25.0 - 90.0 °C = -65.0

⇒ mass of the water = 25.0 grams

⇒ c(water) = the specific heat of water = 4.18 J/g°C

⇒ ΔT(water) = T2 - T1 = 25.0 - 22.5 = 2.5°C

12.0 * c(metal) * -65.0 = -25.0 * 4.18 * 2.5

c(metal) = 0.335 J/g°C

The specific heat of the metal is 0.335 J/g°C