A 150.0 g sample of a metal at 75.0 o C is added to 150.0 g of H2O at 15.0 o C. The temperature of the water rises to 18.3 o C. Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water. The specific heat of water is 4.186 joule/gram-°C.

The specific heat of the metal is 0.244 J/g°C

Explanation:

Step 1: Data given

Mass of metal = 150.0 grams

Temperature of the metal = 75.0 °C

Mass of water = 150.0 grams

Temperature of the water = 15.0 °C

Final temperature of the water = 18.3 °C

The specific heat of water is 4.184 J/g°C

Step 2: Calculate the specific heat of metal

Qlost = -Qgained

Q= m*C*ΔT

Qmetal = -Qwater

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

⇒m(metal) = the mass of metal = 150.0 grams

⇒c(metal) = the specific heat of the metal = TO BE DETERMINED

⇒ΔT = The change of temperature = 18.3 - 75.0 = -56.7 °C

⇒m(water) = the mass of water = 150.0 grams

⇒c(water) = the specific heat of water = 4.186 J/g°C

⇒ΔT(water) = T2 - T1 = 18.3 - 15.0 = 3.3 °C

150.0 * c(metal) * -56.7 = -150.0 * 4.186 * 3.3

c(metal) = 0.244 J/g°C

The specific heat of the metal is 0.244 J/g°C