Question

A 234.0 g piece of lead is heated to 86.0oC and then dropped into a calorimeter containing 611.0 g of water that initally is at 24.0oC. Neglecting the heat capacity of the container, find the final equilibrium temperature (in oC) of the lead and water.

Answer 1

Answer:
`24.70 ^(\circ)C`

Step-by-step explanation:

Given

mass of lead piece
`m_l=234 gm\approx 0.234 kg`

mass of water in calorimeter
`m_w=611 gm\approx 0.611 kg`

Initial temperature of water
`T_w=24^(\circ)C`

Initial temperature of lead piece
`T_l=24^(\circ)C`

we know heat capacity of lead and water are
`125.604 J/kg-k` and
`4.184 kJ/kg-k` respectively

Let us take
`T ^(\circ)C` be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

`m_lc_l(T_l-T)=m_wc_w(T-T_w)`

`0.234* 125.604(86-T)=0.611* 4.184* 1000(T-24)`

`86-T=(0.611* 4.184* 1000)/(29.391)(T-24)`

`86-T=86.97T-2087.49`

`T=(2173.491)/(87.97)=24.70^(\circ)C`