A 0.065 kg ingot of metal is heated to 210◦C and then is dropped into a beaker containing 0.377 kg of water initially at 26◦C. If the final equilibrium state of the mixed system is 28.4 ◦C, find the specific heat of the metal. The specific heat of water is 4186 J/kg · ◦ C. Answer in units of J/kg · ◦ C.

Explanation:

Given that,

Metal of mass

M = 0.065kg

Initial temperature of metal

θm = 210°C

The metal is drop into a beaker which contain liquid of mass

m = 0.377 kg

Initial temperature of water

θw = 26°C

The final mixture temperature is

θf = 28.14°C

Specific heat capacity of water

Cw = 4186 J/kg°C

Since the metal is hotter than the water, then the metal will lose heat, while the water will gain heat, we assume that no heat is loss by the beaker.

So,

Heat Loss = Heat gain

Now, heat loss by metal

H(loss) = M•Cm•∆θ

Where M is mass of meta

Cm is specific capacity of metal, which we are looking fro

So,

H(loss) = 0.065 × Cm × (θi - θf)

H(loss) = 0.065 × Cm × (210-28.14)

H(loss) = 11.821 •Cm

Now, Heat gain by water

H(gain) = m•Cw•∆θ

H(gain) = m•Cw•(θf - θi)

Where

m is mass of water and Cw is specific heat capacity of water

H(gain) = 0.377 ×4286 × (28.14-26)

H(gain) = 3457.86

So, H(loss) = Heat(gain)

11.821 •Cm = 3457.86

Cm = 3457.86/11.821

Cm = 292.52 J/Kg°C

The specific heat capacity of the metal ball is 292.52 J/Kg°C