Question

Question in the details please help asap would be greatly appreciated?

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

Answer 1

Well let us discuss first what is happening here.

When you heat the lead ball it receives heat energy and its temperature increases gradually and at a time reaches its melting point.

Then it would require certain amount of energy,which will be required to melt it i.e the latent heat.

Now,let,the lead ball was initially at a temperature of `theta`,so to take its temperature to its melting point,heat energy required will be `m×s×((330+273) - theta)`(where, `m` is its mass and `s` is specific heat)

So, this value becomes `1.45×120×(603-theta) J=H'`(let)

Now,see the unit of latent heat,it's `J/(kg)` that means `2.45×10^4 J` of latent heat is required per kg of lead to change its state from solid to liquid.

So,for `1.45 Kg` of lead,latent heat required will be `(1.45×2.45×10^4)J=3.54×10^4J=H`(let)

So,for this entire process heat energy required will be `H'+H`

Now given, `H+H' =4.46×10^4`

So,you can solve the equation and you get `theta = 550.13 K` or, `(550.13-273) = 277.13 @^C`