Question

In: Physics

Suppose 28.0 g of ice at -10.0∘C is placed into 300.0 g of water in a...

Suppose 28.0 g of ice at -10.0∘C is placed into 300.0 g of water in a 200.0-g copper calorimeter. The final temperature of the water and copper calorimeter is 18.0∘C.

1) What was the initial common temperature of the water and copper? (Express your answer to three significant figures.)

Expert Solution

Specific heat of water = C₁ = 4186 J/(kg.^oC)

Specific heat of copper = C₂ = 385 J/(kg.^oC)

Specific heat of ice = C₃ = 2100 J/(kg.^oC)

Latent heat of fusion of water = L = 3.34 x 10⁵ J/kg

Mass of the water = m₁ = 300 g = 0.3 kg

Mass of the copper calorimeter = m₂ = 200 g = 0.2 kg

Mass of the ice = m₃ = 28 g = 0.028 kg

Initial temperature of the water and copper calorimeter = T₁

Initial temperature of ice = T₂ = -10 ^oC

Melting point of ice = T₃ = 0 ^oC

Final equilibrium temperature = T₄ = 18 ^oC

Heat removed from the water to take it to 18 ^oC = Q₁

Q₁ = m₁C₁(T₁ - T₄)

Heat removed from the copper calorimeter to take it to 18 ^oC = Q₂

Q₂ = m₂C₂(T₁ - T₄)

Heat added to the ice to take it to 18 ^oC = Q₃

Q₃ = m₃C₃(T₃ - T₂) + m₃L + m₃C₁(T₄ - T₃)

The heat removed from the water and the copper calorimeter is equal to the heat added to the ice.

Q₁ + Q₂ = Q₃

m₁C₁(T₁ - T₄) + m₂C₂(T₁ - T₄) = m₃C₃(T₃ - T₂) + m₃L + m₃C₁(T₄ - T₃)

(0.3)(4186)(T₁ - 18) + (0.2)(385)(T₁ - 18) = (0.028)(2100)(0 - (-10)) + (0.028)(3.34x10⁵) + (0.028)(4186)(18 - 0)

1255.8T₁ - 22604.4 + 77T₁ - 1386 = 588 + 9352 + 2109.744

1332.8T₁ = 36040.144

T₁ = 27.0 ^oC

1) Initial common temperature of the water and copper calorimeter = 27.0 ^oC

genius_generous answered 2 years ago

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