Question

A 30.0 g sample of a metal was heated in a hot water bath to 80°c. it was then quickly transferred to a coffee-cup calorimeter. the calorimeter contained 100.0 g of water at a temperature of 20°c. the final temperature of the contents of the calorimeter was 25°c. what is the specific heat capacity of the metal?

Answer 1

Answer:  0.303 cal/g°C

Explanation:

1) Data:

a) Metal:

m₁ = 30.0 g

T₁ =  80°C

Cs₁ = ?

b) Coffee-cup calorimeter (water).

m₂ = 100.0 g

T₂ = 20°C

Cs₂ = 1 cal / g°C (from tables or internet)

c) Mixture.

Tf = 25°C.

2) Physical principles and formulae:

• Since the calorimeter is a closed insulated system, heat released by the metal = heat absorbed by the water: Q water = Q metal

• Q = m × Cs × ΔT

3) Solution:

• Q metal = Q water ⇒ m₁Cs₁ΔT₁ = m₂Cs₂ΔT₂

• ∴ 30.0g (Cs₂) (80°C - 25°C) = 100.0g (1 cal/g°C) (25°C - 20°C)

• ⇒ Cs₂ = [ 100.0 (5) / 30(55) ] cal/g°C = 0.303 cal/g°C

Answer 2

when the metal  is lost heat and the calorimeter of water is gained the heat 

and when the heat lost = the heat gained so,

(M*C*ΔT)m =  (M*C*ΔT)w

when Mm= mass of the metal = 30 g 

Δ Tm = (80-25) = 55 °C

and Mw = mass of water = 100 g  

Cw is the specific heat of water = 4.181 J/g.°C

ΔTw = (25-20) = 5 °C

so by substitution:

∴ 30* Cm*55 = 100 * 4.181 * 5 

∴Cm (specific heat of metal) = (100*4.181*5)/(30*55) 

∴C of metal = 1.267 J/g.°C