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.

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

mavila18 mavila18 · Answer 2 · 2020-04-29T04:53:08+02:00

Answer:

320.86J/kgC

Explanation:

To find the specific heat of the substance you take into account that the heat lost by the metal is gained by the water, that is:

[tex]Q_1=-Q_2[/tex]

Furthermore the heat is given by:

[tex]Q_1=m_1c(T_1-T)\\\\Q_2=m_2c(T_2-T)[/tex]

m1: mass of the metal

m2: mass of the water

c: specific heat

T: equilibrium temperature

T1: temperature of the metal

T2: temperature of water

By replacing all these values you can calculate c of the metal:

[tex]m_1c_1(T_1-T)=-m_2c_2(T_2-T)\\\\c_1(0.065kg)(210-28.4)\°C=-(0.377kg)(4186J/kg\°C)(26-28.4)\°C\\\\c_1=320.86\frac{J}{kg\°C}[/tex]

Hence, the specific heat of the metal is 320.86J/kgC