If 50,000joules of heat is added to 10kg of lead at its melting point of 330 degrees celsius.If 2kg of lead melts,calculate the latent heat of fusion of lead
It took 50,000 joules to melt 2 kg of lead so
25,000 Joules/kg
To calculate the latent heat of fusion of lead, we need to use the formula:
Q = m * L
where Q is the amount of heat added, m is the mass of the substance, and L is the latent heat of fusion.
Let's break down the problem step by step:
Step 1: Calculate the heat added to melt 2 kg of lead
Since 10 kg of lead was initially at its melting point and 2 kg melted, the remaining 8 kg did not melt and remained solid. Therefore, the heat added to melt 2 kg of lead is:
Q = m * c * ΔT
where Q is the heat added, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.
Given that the specific heat capacity of lead is 130 J/(kg·°C) and the melting point of lead is 330°C, we can calculate the heat added:
Q = 2 kg * 130 J/(kg·°C) * (330°C - 0°C)
Q = 85800 J
Step 2: Calculate the latent heat of fusion
Now that we have the heat added to melt 2 kg of lead (Q = 85800 J), we can use it to calculate the latent heat of fusion (L):
Q = m * L
85800 J = 2 kg * L
Rearranging the equation, we find:
L = Q / m
L = 85800 J / 2 kg
L = 42900 J/kg
Therefore, the latent heat of fusion of lead is 42900 J/kg.