A 110.0-g sample of metal at 82.00°C is added to 110.0 g of H2O(l) at 27.00°C in an insulated container. The temperature rises to 30.56°C. Neglecting the heat capacity of the container, what is the specific heat of the metal? The specific heat of H2O(l) is 4.18 J/(g ∙ °C). Group of answer choices 4.18 J/(g ∙ °C) 60.4 J/(g ∙ °C) 0.289 J/(g ∙ °C) 0.289 J/(g ∙ °C) 14.4 J/(g ∙ °C)

The specific heat of the metal is 0.289 J/g°C

Explanation:

Step 1: Data given

Mass of metal = 110.0 grams

Temperature of the metal = 82.00 °C

MAss of water = 110.0 grams

Temperature of the water = 27.00 °C

The final temperature = 30.56 °C

The specific heat of H2O is 4.18 J/(g°C).

Step 2: Calculate the specific heat of the metal

Heat lost = heat gained

Qlost = - Qgained

Qmetal = - Qwater

Q =m*c*ΔT

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

⇒with m(metal) = the mass of the metal = 110.0 grams

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

⇒with ΔT(metal) = the change of temperature of the metal = T2 - T1 = 30.56 - 82.00 °C= -51.44 °C

⇒with m(water) = the mass of water = 110.0 grams

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

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = 30.56 °C - 27.00 °C = 3.56 °C

110.0 * c(metal) * -51.44 = -110.0 * 4.18 * 3.56

c(metal) = 0.289 J/g°C

The specific heat of the metal is 0.289 J/g°C

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