Question

A student obtained an unknown metal sample that weighed 65.3 g and at a temperature of 99.8^oC, he placed it in a calorimeter containing 43.7 g of water at 25.7^oC. At equilibrium the temperature of the water and metal was 34.5^oC.

    Knowing the specific heat of the water to be 4.18 J/g^oC, what is the specific heat of the metal?

[ Hint: S.H._water x m_water x Δt _water= S.H._metal x m_metal x Δt _metal ]

Answer 1

Given,

Mass of the metal sample = 65.3 g

The initial temperature of the metal sample(T_i) = 99.8 ^oC

The final temperature of the water and metal at equilibrium (T_f) = 34.5 ^oC

Mass of water in a calorimeter = 43.7 g

The initial temperature of water(T_i) = 25.7 ^oC

the specific heat of water(C_w) = 4.18 J/ g ^oC

Now, when the metal sample is placed in a calorimeter, the equilibrium is established between water and metal, thus,

Heat absorbed by water = Heat lost by the metal

m_water x C_water x T = -( m_metal x C_metal x T)

Here, T = T_f - T_i

43.7 g x 4.18 J/g ^oC x (34.5-25.7) ^oC = - (65.3 g x C_metal x (34.5 -99.8)^oC

1607.461 = - [-4264.09 x C_metal ]

1607.461 = [4264.09 x C_metal ]

C_metal = 1607.461/4264.09

C_metal = 0.377 J/g ^oC

Thus, specific heat of metal = 0.377 J/g ^oC[3 S.F]