Question

a piece of metal with a mass of 15.3 grams has a temperature of 50.0°C. When the metal is placed in 80.2 grams of water at 21.0°C, the temperature rises by 4.3°C. What is the specific heat capacity of the metal?

Answer 1

1.21

Step-by-step explanation:

Heat rise in the body happens due to heat supplied by water to the body.

Heat rise in body = m₁ c₁ ΔT₁

Where m₁ is mass of body and c₁ is its specific heat of body

Heat lost from water to the body = m₂ c₂ ΔT₂

Where m₂ is mass of water and c₂ is its specific heat of water ( c₂ =1 (since water))

Equating both:

15.3 x c₁ x 4.3 = 80.2 x 1 x 4.3

⇒ c₁ = 80.2 / (15.3 x 4.3) = 1.21

Answer 2

c₁ = 3.68 J/g°C

Step-by-step explanation:

The quantity of heat Q gained by a body of mass m, specific heat capacity, c and temperature change ΔT is Q = mcΔT. Let m₁, c₁ and m₂, c₂ represent the masses and specific heat capacities of the metal and water respectively. ΔT₂ for the water = 4.3°C. This means that its final temperature of the water after the metal has been placed in it is 21.0°C + 4.3°C = 24.3°C, which is the final temperature of the metal. So ΔT₁ for metal = 24.3°C - 50.0°C = -25.7°C.

Since heat loss equals heat gain

-m₁c₁ΔT₁ = m₂c₂ΔT₂

c₁ = -m₂c₂ΔT₂/m₁ΔT₁

c₁ = -80.2 × 4.2 × 4.3/15.3 × -25.7 = -1448.412/(-393.21) = 3.683 J/g°C

c₁ ≈ 3.68 J/g°C