The temperature of a brass cylinder of mass 200 gram was raised 1 hundred degree Celsius and transferred to a thin aluminium can of negligible heat capacity containing 150 gram of paraffin at 11 degree Celsius.If the final study temperature after studying was 20 degree C. Calculate the specific latent heat of paraffin (neglect heat losses specific heat capacity of brass is 380 J g-1 degree C-1.

Dear Student,
Specific latent heat can't be determined in this problem with the given data. What can be determined is heat capacity of Paraffin. The calculation is as follows:

Mass of the brass cylinder, M = 200 g 
Temperature of the brass cylinder, ${\theta }_1={100}^{o}C$
Mass of parafin wax, m = 150 g 
Temperature of parafin wax, ${\theta }_2={11}^{o}C$
When transferred to aluminium can, total mass = M + m = 0.200 kg + 0.150 kg = 0.350 kg
Final temperature, $\theta ={20}^{o}C$
The amount of heat lost by brass cylinder when transferred to paraffin $=Mc\left({\theta }_1-\theta \right)=200\times 380\times \left(100-20\right)=6.08 \times {10}^6 J$
The amount of heat gained by the Paraffin wax = $m\times heat capacity\times \left(\theta -{\theta }_2\right)=150\times heat capacity\times \left(20-11\right)=1350\times heat capacity$
Given, no heat losses, so
heat lost by brass cylinder = heat gained by the Paraffin wax
$$\begin{gathered} 6.08\times {10}^6=1350\times heat capacity \\ heat capacity=\frac{6.08\times {10}^6}{1350}=4503.7 J{g}^{-1} C^{-1} \end{gathered}$$
Regards