Ahot (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

i think the answer would be reserve.