Answer:
[tex]1.2\cdot 10^5 J[/tex]
Explanation:
1) First of all, we need to calculate the heat needed to raise the temperature of the block of lead from the initial temperature (25.0 C) to the melting temperature (327.5 C). This heat is given by the equation
[tex]Q_1 = m C_{pb} \Delta T[/tex]
where
m = 2.00 kg is the mass of the block
Cpb = 130.0 J/kg*C is the specific heat capacity of lead
[tex]\Delta T=327.5 C-25.0 C=302.5 C[/tex] is the temperature difference
Substituting,
[tex]Q_1 = (2.00 kg) (130 J/kgC) (302.5 C)=7.87\cdot 10^4 J[/tex]
2) Now we have to calculate the amount of heat needed to completely melt the lead, which is given by:
[tex]Q_2 = m Hj[/tex]
where
Hj = 2.04x10^4 J/kg is the latent heat of fusion of lead
Substituting,
[tex]Q_2 = (2.00 kg)(2.04\cdot 10^4 J/kg)=4.08\cdot 10^4 J[/tex]
3) Therefore, the total heat needed is
[tex]Q=Q_1+Q_2=7.87\cdot 10^4 J+4.08\cdot 10^4 J=1.2 \cdot 10^5 J[/tex]
Answer:
Q = 2.00kg x 130 J/kgC x 302.5C
Q = 7.87 x 10^4j
Q = 2.00kg x 2.04 x 20^4
Q= 4.08 x 10^4
Q = 7.87 x 10^4j + 4.08 x 10^4 = 1.2 x 10^5j
Explanation:
