Question

A 151.5-g sample of a metal at 75.0°C is added to 151.5 g at 15.1°C. The temperature of the water rises to 18.7°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 capacity of water is 4.18 J/°C·g. Specific heat capacity = J/°C·g Submit AnswerTry Another Version

Answer 1

Answer:

The specific heat capacity of the metal is 0.268 J/g°C

Explanation:

Step 1: Data given

Mass of the metal = 151.5 grams

The temperature of the metal = 75.0 °C

Temperature of water = 15.1 °C

The temperature of the water rises to 18.7°C.

The specific heat capacity of water is 4.18 J/°C*g

Step 2: Calculate the specific heat capacity of the metal

heat lost = heat gained

Q = m*c*ΔT

Qmetal = - Qwater

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

⇒ mass of the metal = 151.5 grams

⇒ c(metal) = TO BE DETERMINED

⇒ΔT( metal) = T2 - T1 = 18.7 °C - 75.0 °C = -56.3 °C

⇒ mass of the water = 151.5 grams

⇒ c(water) = 4.184 J/g°C

⇒ ΔT(water) = 18.7° - 15.1 = 3.6 °C

151.5g * c(metal) * -56.3°C = 151.5g * 4.184 J/g°C * 3.6 °C

c(metal) = 0.268 J/g°C

The specific heat capacity of the metal is 0.268 J/g°C