Question

One kg of lead at its melting point $327°C$ is dropped in ${1\,kg}$ of water at $20°C$. Assuming no heat is lost in the surrounding. Calculate final temperature of water (1 specific heat capacity of lead=$130 J/ Kg K$ , specific heat capacity of water= $4200 J/Kg K$ and specific heat of fusion of lead= $27000 J/Kg$)

Answer 1

Hint: Let us first understand what is meant by melting point and specific heat. The temperature at which the substance converts from solid state to liquid state is known as melting point. The amount of heat required to rise the temperature of a substance by one degree celsius per unit mass of a substance.

Complete step by step answer:
According to the principle of calorimetry,the total heat lost by the hot body is equal to the total heat gained by the cold body. This principle indicates the law of conservation of energy.
Given;
Specific heat capacity of lead , ${C_lead}= 130 J/Kg K$
Specific heat capacity of water,${C_water}= 4200 J/Kg K$
Specific heat of fusion of lead, $L= 27000 J/Kg$
${m_1} =1 kg$ of lead
${m_2} =1 kg$ of water
Assume the final temperature of the mixture be t.
Heat lost by Lead = Heat gained by Water
\[
{m_1}L + {m_1} \times {C_{lead}}(327 - t) = {m_2} \times {C_{water}}(t - 20) \\
\Rightarrow 1 \times 27000 + 1 \times 130(327 - t) = 1 \times 4200(t - 20) \\
\Rightarrow27000 + 42510 - 130 t = 4200 t - 84000 \\
\Rightarrow 153510 = 4330 t \\
\Rightarrow t = \dfrac{{153510}}{{4330}} \\
\therefore t = 35.45^\circ C \\
 \]
Hence,the final temperature of the water is \[ 35.45^\circ C\].

Note:Do not get confused between latent heat and specific heat. Latent heat is the energy required by a substance to undergo a phase change. Specific heat is the amount of heat required by one gram of a substance to rise the temperature by one degree celsius. Remember that specific is not applicable when a substance undergoes a phase change.