Question

a piece of aluminium of mass 0.5kg is heated to 100°C and then placed in 0.4kg of water at 10°C.If the resulting temperature of the mixture is 30°C, what is the specific heat capacity of aluminium

Answer 1

Answer:

$c_{Al}=960\:J/Kg^0C$.

Step by step explanation:

The givens are:

• Mass of aluminium: $m_{Al}=0.5\:kg$
• Temperature of aluminium: $T_{Al}=100^0C$
• Mass or water: $m_w=0.4\:kg$
• Temperature of water: $T_w=10^0C$
• Temperature of mixture: $x=30^0C$

Heat gained by water = Heat lost by aluminium

$m_wc_w(x-T_w)=m_{Al}c_{Al}(T_{Al}-x)$; where $c_w$ is the specific heat capacity of water = $4200\:J/Kg^0C$

$0.4*4200*(30-10)=0.5*c_{Al}*(100-30)\\c_{Al}=960\:J/Kg^0C$.