Question

A lead metal ball of mass 80 g is fired from the canon at temperature of 87 C and hits a wall made of steel and melts down. Find the minimum speed of the lead ball must have? (Latent heat of melting of lead 24.5x10' J/kg)​

Answer 1

Given that,

Initial temperature = 87°C

Mass = 80 g

Latent heat of melting of lead $L=2.45\times10^{4}\ J/kg$

We know that,

Melting temperature of lead = 327.5°C

The heat energy need to melt metal ball is

$Q=Q_{1}+Q_{2}$

We need to calculate the heat energy

Using formula of heat energy

$Q_{1}=ms\Delta T$

Where, T = temperature

sw = specific heat

We need to calculate the heat energy

Using formula of heat energy

$Q_{2}=mL$

Where, L = Latent heat

m = mass

We need to calculate the minimum speed of the lead ball

Using conservation of energy

$K.E=Q$

$K.E=Q_{1}+Q_{2}$

$\dfrac{1}{2}mv^2=mL+ms\Delta T$

$v^2=2(L+s\Delta T)$

Put the value into the formula

$v^2=2(2.45\times10^{-4}+125\times(327.5-87))$

$v=\sqrt{2(2.45\times10^{-4}+125\times(327.5-87))}$

$v=245.20\ m/s$

$v=882.72\ km/hr$

Hence, The minimum speed of the lead ball is 882.72 km/hr.