# Question 7ca3b

Mar 1, 2017

Let $s$ be specific heat capacity of the unknown substance.
Heat Lost by this substance when put in calorimeter is found out the following expression
$Q = m s \Delta t$ .....(1)
${Q}_{\text{lost"=5.00xxsxx(80.0-45.0)=175s" J}}$ .......(2)

Similarly using (1) to find out Heat gained by water and calorimeter combination
${Q}_{\text{gained"=Q_"water"+Q_"calorimeter}}$
${Q}_{\text{gained}} = 30.0 \times 4.1813 \times \left(45 - 20\right) + 10.0 \times \left(45 - 20\right)$
${Q}_{\text{gained"=3136+250=3386" J}}$ .....(3)

Using Law of conservation of energy and equating (2) with (3)
${Q}_{\text{gained"=Q_"lost}}$
$3386 = 175 s$
=>s=19.35" J(g"^@"C")^-1#