Question

When \(M_1~\text{gram}\) of ice at \(–10^\circ C\) (specific heat \(s=0.5 cal/g^\circ C)\) is added to \(M_2~\text{gram}\) of water at \(50^\circ C\), finally no ice is left and the water is at \(0^\circ C\). The value of latent heat of ice, in \(cal/g\) is

Answer 1

Given: \(\theta_1=-10^\circ C, s_{ice}=0.5~cal/g^\circ C, \theta_2=50^\circ C, \theta=0^\circ C\)
\(\text{Heat loss} = \text{Heat gain}\)
\(M_2s_w(50-0)=M_1s_{ice}(0-(-10))+M_1L_f\)
\(M_2\times1\times50=M_1\times0.5\times10+M_1\times L_f\)
\(L_f=\dfrac{50M_2-5M_1}{M_1}\)
\(=\dfrac{50M_2}{M_1}-5\)

Final Answer: (c)