Question

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

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

Answer 1

Answer: (C)

In cooling $200g$ 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
$Q_2=m_{ice}{s}_{\text{ice }}(0-(-14)+m_{\text{ice }}{L}_{f_{\text{ice }}}+m_{\text{ice }}{s}_{\text{water }}(10-0)$
$=m_{\text{ice }}\times 0.5\times 14+m_{\text{ice }}\times 80+m_{\text{ice }}\times 1\times 10=97m_{\text{ice }}$
As $Q_1=Q_2$
$\therefore 97m_{\text{ice }}=3000$
or $m_{\text{ice }}=31g$