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

Answer:

The final temperature is  $31.3^{\circ}\, C$

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 $m_{al}C_{al}(90-T)=m_wC_w(T-22)+m_{cu}C_{cu}(T-22)$

=>$300\times 0.9\times (90-T)=(400\times 4.186+80\times 0.386)(T-22)$

=>$T=31.3^{\circ}\, C$

Thus the final temperature is  $31.3^{\circ}\, C$