Question

What would be the final temperature if a 2 kg piece of lead at 200 °C is insertedin a container with 10 kg of water at 50 °C ? (specific heat capacity of lead, c'=128 J kg-1 °C !)?

Answer 1

Given that the mass of lead, m = 2kg and initial temperature, Tl= 200 deg C

and the mass of water, M = 10kg and initial temperature of water be Tw = 50 deg C

Also, the heat capacity of lead is, Cl = 128 J /kg deg C

and heat capacity of water is Cw = 4184 J /kg deg C

We have to find the final temperature of the lead and water.

As lead is immersed in water, temperature exchange occurs between water and lead until the temperature of lead and water becomes the same. this will be the final temperature, let this final temperature be denoted by T.

So,

`\begin{gathered} \text{Heat of lead = Heat of water } \\ \end{gathered}`

Let the heat of lead be Ql and the heat of water be Qw.

`mC_l(T-T_l)=MC_w(T-T_w)`

Substituting the values in the above equation,

`\begin{gathered} 2*128*(T-200)=10*4184*(T-50) \\ 41840T-256T=\text{ 209200-51200} \\ T=(2040800)/(41584) \\ \\ =49.07^(\circ)C \end{gathered}`