This question is bit complicated. We need to understand this step by step.

• First we have a sample of water (0.8kg) at 15C in a calorimeter.
• Then molten lead at T temperature is added in to the calorimeter.
• T>melting point of lead. So initially lead drops it temperature while it comes to melting point.
• When lead comes to melting point it uses its latent heat of fusion and become lead solid.
• When it become solid it will further decrease its temperature and the system will come to equilibrium
• At all these time water and calorimeter will absorb all the energy released by lead.

Let us first consider energy released by Lead

Heat released to become lead at melting point = Q1

Heat released at latent heat of fusion = Q2

Heat released up to equilibrium  = Q3

`Q = mCtheta`

`Q1 = 0.4xx157xx(327-T)`

`Q = mL`

`Q2 = -0.4xx2.31xx10^4 = -4000J = -9.24KJ`

`Q = mcCtheta`

`Q3 = -0.4xx136xx(25-327) = -16428.8 = -16.43KJ`

(-) sign is used because the energy is released.

For absorbing energy

Energy absorbed by water = Q4

Energy absorbed by calorimeter = Q5

`Q = mCtheta`

`Q4 = 0.8xx4182xx(25-15) = 33456J = 33.46KJ`

`Q5 = Ctheta`

`Q5 = 42.7xx10 = 427J = 0.427KJ`

Energy released = energy absorbed

`-0.4xx157xx(327-T)/1000+9.24+16.43 = 33.46+0.427`

`T = 457.84`

So the initial temperature of lead is 454.84C

Assumption

There is no heat losses in the system other than the releasing and absorbing heat between water,lead and calorimeter.