Question

Problem 1.4 300 g of aluminum at 90°C is placed in a calorimeter cup containing 400 g of water. The mass of the copper calorimeter cup 80 g. The initial temperature of the water and cup is 22.0°C. What is the final temperature?

Answer 1

The final temperature is
`31.3^(\circ)\, C`

Step-by-step explanation:

Specific heat capacity of aluminum ,
`C_(al)=0.900 J/gmK`

Specific heat capacity of water,
`C_w=4.186 J/gmK`

Specific heat capacity of copper ,
`C_(cu)=0.386 J/gmK`

Let the final temperature be T degree C

Since Heat lost by aluminium = Heat gained by water and calorimeter

Therefore
`m_(al)C_(al)(90-T)=m_wC_w(T-22)+m_(cu)C_(cu)(T-22)`

=>
`300* 0.9* (90-T)=(400* 4.186+80* 0.386)(T-22)`

=>
`T=31.3^(\circ)\, C`

Thus the final temperature is
`31.3^(\circ)\, C`