23 kJ
given:
\(m _{\text {ice }}=50 g =0.05 kg\)
\(l_{\text {ice }}=3.34 \times 10^{5} Jkg ^{-1}\)
\(c_{\text {ice }}=2100 Jkg ^{-} k ^{-1}\)
\(c_{ w }=4200 Jkg ^{-1}\)
it is important, note that the required process involves
three stages:
Stage 1
warming up of icc \(-5^{\circ} C\) to ice at \(0^{\circ} C\)
i.e. ice at \(-5^{\circ} C\) wi ns up to ice at \(0^{\circ} C\) heat \(q _{1}= m _{\text {cec }} c _{\text {ice }} \Delta \theta\)
\(q_{1}=0.05 \times 2100 \quad[0-(-10)]\)
\(=0.05 \times 2100 \cdot 10\)
\(q_{1}=1050 J\)
Stage 2 - melting of ice at \(0^{\circ} C\) to water at \(0^{\circ} C\) (change of statc)
heat \(q_{2}=m l_{ f }\)
\(q_{2}=0.05 \times 3.34 \times 10^{5}\)
\(q_{2}=16700 J\)
Stage 3
warming up water at \(0^{\circ} C\) to water at \(25^{\circ} C\)
\(\left( NB m_{\text {ice }}=m_{\text {water }}\right)\)
heat \(q_{3}=m_{ w } c_{ w } \Delta \theta\)
\(=0.05 \times 4200(25-0)\)
\(=0.05 \times 4200 \times 25\)
\(=5,250 J\)
\(\therefore\) Total heat consumed
\(Q=7_{1}+q_{2}+q_{3}\)
\(=1950 J +16700 J +5,250 J\)
\(=23,000 J =23 kJ\)