A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimeter containing 75 g of water at 20°C.

The calorimeter is constructed of a material that has a specific heat of 0.10 cal / g⋅°C.

When equilibrium is reached, what will be the final temperature? cwater = 1.00 cal/g⋅°C.

a. 114°C

b. 72°C

c. 64°C

d. 37°C

d. 37°C

Explanation:

When Equilibrium is reached,

Heat lost = heat gain.

Heat lost by the hot metal = c₁m₁ (T₁-T₂) ... equation 1

Where c₁ = specific heat capacity of the metal = 0.25 cal/g⋅°C, m₁ = mass of the metal = 250 g, T₁ = initial Temperature of the metal = 70°C, T₂ final temperature of the metal.

Substituting this values into equation 1,

Heat lost by the metal = 62.5 (70-T₂)

Also,

Heat gain by the water = c₂m₂ (T₂-T₁) ... equation 2

c₂ = 1.00 cal/g⋅°C., m₂ = 75 g, T₁ = 20°C

Substituting this values into equation 2,

Heat gain by water = 1 * 75 (T₂ - 20)

Heat gained by water = 75 (T₂ - 20)

Also,

Heat gained by the calorimeter = c₃m₃ (T₂-T₁) ... equation 3

Where c₃ = 0.10 cal / g⋅°C, m = 500 g, T₁ = 20°C.

Substituting this values into equation 3

Heat gained by the calorimeter = 0.1 * 500 (T₂ - 20)

Heat gained by the calorimeter = 50 (T₂ - 20)

Heat lost by the metal = heat gained by water + heat gained by the calorimeter.

62.5 (70-T₂) = 50 (T₂ - 20) + 75 (T₂ - 20)

4375 - 62.5T₂ = 50T₂ - 1000 + 75T₂ - 1500

Collecting like terms,

-62.5T₂ - 50T₂ - 75T₂ = - 1000 - 1500 - 4375

-187.5T₂ = - 6875

Dividing both side by the coefficient of T₂

-187.5T₂ / - 187.5 = - 6875 / -187.5

T₂ = 36.666°C

T₂ ≈ 37°C

The final temperature = 37°C