A sample of lead with a mass of 1.45kg is heated until it reaches its melting point at 330C. The lead is then heated further until it has entirely melted. During this process the lead absorbs a total of 4.46 104J as heat. What is the temperature at which the lead is initially heated? The specific heat capacity of solid lead over the range of temperatures in question is 120J/kg, and the latent heat of fusion for lead is 2.45 104J/kg

please help its really confusing

Initial heat =1.45*c*(330-Ti)

c=120J/kg
final heat=initial heat+2.45e4+1.45*c*(330-Ti)

solve for Ti

To solve this problem, we need to consider the energy absorbed by the lead during the two stages: heating to the melting point and then melting.

First, let's calculate the energy required to heat the solid lead to its melting point:
Q1 = m * c * ΔT

Where:
Q1 = Energy absorbed (4.46 * 10^4 J)
m = Mass of the lead (1.45 kg)
c = Specific heat capacity of solid lead (120 J/kg°C)
ΔT = Change in temperature (final temperature - initial temperature)

We need to find the initial temperature (T_initial), so we rearrange the equation to solve for ΔT:
ΔT = Q1 / (m * c)

Substitute the given values:
ΔT = (4.46 * 10^4 J) / (1.45 kg * 120 J/kg°C)

Now, we can calculate the change in temperature:
ΔT = 31.04°C

Since we know the lead reaches its melting point at 330°C, we can find the initial temperature:
T_initial = T_melting - ΔT
T_initial = 330°C - 31.04°C
T_initial = 298.96°C

Therefore, the initial temperature at which the lead is heated is 298.96°C.