Question

How many gram of ice at $-14^{\circ} C$ is needed to cool $200 \,g$ of water from $25^{\circ} C$ to $10^{\circ} C ?$ (Specific heat of ice $=0.5\, cal / g ^{\circ} C$ and latent heat of fusion of ice $=80\, cal / g )$

• A. 11 g
• B. 21 g
• C. 31 g
• D. 41 g

Answer 1

Answer: (C)

In cooling $200\, g$ of water from $25^{\circ} C$ to $10^{\circ} C$ heat to be extracted from water is
$Q_{1}=m_{\text {water }} s_{\text {water }}(25-10)$
$=200 \times 1 \times(25-10)=3000$ cal
Heat absorbed by $m_{\text {ice }} g$ ice at $-14^{\circ} C$ to convert it to water at $10^{\circ} C$ is
$Q_{2} =m_{ ice } s_{\text {ice }}\left(0-(-14)+m_{\text {ice }} L_{f_{\text {ice }}}+m_{\text {ice }} s_{\text {water }}(10-0)\right.$
$=m_{\text {ice }} \times 0.5 \times 14+m_{\text {ice }} \times 80+m_{\text {ice }} \times 1 \times 10=97\, m_{\text {ice }}$
As $Q_{1}=Q_{2}$
$\therefore 97\, m_{\text {ice }}=3000$
or $m_{\text {ice }}=31 \,g$