Question

An amateur blacksmith wants to cool off a 42kg glowing piece of iron, specific heat 470and decides to toss it into a 5.0 kg iron bucket with 10.0 kg of room temperature(23 C) water in it. To his surprise the water completely evaporates away (meaning once it vaporizes it is no longer part of the system) and after some time he goes to pick up the bucket but finds that the bucket is at 150 C.

Required:
What was the initial temperature of the glowing piece of iron?

Answer 1

Answer:

The right approach is "1479°C".

Explanation:

The given values are:

Mass of iron piece,

$m_p=42 \ kg$

Mass of iron bucket,

$m_I=5 \ kg$

Mass of water,

$m_w=10 \ kg$

Iron's specific heat,

$C_I=470 \ J/Kg^{\circ}C$

Water's specific heat,

$C_w=4186 \ J/Kg^{\circ}C$

Initial temperature,

$t_I=23^{\circ}C$

Final equilibrium temperature,

$T=150^{\circ}C$

Latent heat,

$L_v=2260\times 10^3 \ J/Kg$

As we know,

The heat lost by the glowing piece of iron will be equal to the heat gain by the iron bucket as well as water, then

⇒ $m_IC_I \Delta T=m_wC_w(100-23)+m_wL_v+m_bC_I(150-23)$

On substituting the given values, we get

⇒ $42\times 420\times \Delta T=10\times 4186(100-23)+10(2260\times 10^3)+5\times 420(150-23)$

⇒ $17640 \Delta T=3.22\times 10^6+2.26\times 10^7+2.667\times 10^5$

⇒          $\Delta T=\frac{2.60867\times 10^7}{17640}$

⇒          $\Delta T=1479^{\circ}C$