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A new metal alloy is found to have a specific heat capacity of 0.260 J/(g⋅∘C). First, 47 g of the new alloy is heated to 180. ∘C. Then, it is placed in an ideal constant-pressure calorimeter containing 110. g of water (Cs,water=4.184 J/(g⋅∘C)) at an initial temperature of 20.0 ∘C. What will the final temperature of the mixture be after it attains thermal equilibrium?

Answer :

Answer:

The final temperature, at the equilibrium is 24.14 °C

Explanation:

Step 1: Data given

specific heat capacity of alloy = 0.260 J/(g°C)

MAss of alloy = 47 grams

Mass of water = 110 grams

Specific heat of water = 4.184 J/g°C

Initial temperature of water = 20.0 °C

Initial temperature of alloy = 180.0 °C

Step 2: Calculate the final temperature at equilibrium

Heat lost = heat gained

Qlost = -Qgained

Q(alloy) =- Q(water)

Q=m*c*ΔT

Q = m(alloy)*c(alloy)*ΔT(alloy) = -m(water) * c(water)* ΔT(water)

⇒with m(alloy) = the mass of alloy = 47.0 grams

⇒with c(alloy) = the specific heat of alloy = 0.260 J/g°C

⇒with ΔT(alloy) = the change of temperature = T2- T1 = T2 - 180 °C

⇒with m(water) = the mass of water = 110 grams

⇒with c(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature = T2 - 20.0°C

47.0*0.260 * (T2 - 180.0) = - 110 * 4.184 * (T2 - 20.0)

12.22(T2-180.0) = -460.24(T2- 20)

12.22T2 - 2199.6 = -460.24T2 + 9204.8

472.46T2 = 11404.4

T2 = 24.14 °C

The final temperature, at the equilibrium is 24.14 °C

The final temperature, at the equilibrium, is 24.14 °C.

Given:

Specific heat capacity of alloy = 0.260 J/(g°C)

Mass of alloy = 47 grams

Mass of water = 110 grams

Specific heat of water = 4.184 J/g°C

Initial temperature of water = 20.0 °C

Initial temperature of alloy = 180.0 °C

Calculation of final temperature at equilibrium:

Heat lost = heat gained

Qlost = -Qgained

Q(alloy) =- Q(water)

Specific heat can be given as:

Q=m*c*ΔT

Q = m(alloy)*c(alloy)*ΔT(alloy) = -m(water) * c(water)* ΔT(water)

⇒with m(alloy) = the mass of alloy = 47.0 grams

⇒with c(alloy) = the specific heat of alloy = 0.260 J/g°C

⇒with ΔT(alloy) = the change of temperature = T₂- T₁ = T₂ - 180 °C

⇒with m(water) = the mass of water = 110 grams

⇒with c(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature = T₂ - 20.0°C

On subsitting the values:

47.0*0.260 * (T₂ - 180.0) = - 110 * 4.184 * (T₂ - 20.0)

12.22(T₂-180.0) = -460.24(T₂- 20)

12.22T₂ - 2199.6 = -460.24T₂ + 9204.8

472.46T₂ = 11404.4

T₂= 24.14 °C

The final temperature, at the equilibrium, is 24.14 °C

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