2, kg of ice at -20^{circ}C is mixed with 5, | vedclass

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(B) Step $1$: Heat required by $2\, kg$ of ice to reach $0^{\circ}C$ is $Q_1 = m_i \cdot c_{ice} \cdot \Delta T = 2\, kg \cdot 0.5\, kcal/kg/^{\circ}C

Vedclass · Accepted Answer

$6$

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(B) Step $1$: Heat required by $2\, kg$ of ice to reach $0^{\circ}C$ is $Q_1 = m_i \cdot c_{ice} \cdot \Delta T = 2\, kg \cdot 0.5\, kcal/kg/^{\circ}C \cdot 20^{\circ}C = 20\, kcal$.Step $2$: Heat available from $5\, kg$ of water cooling from $20^{\circ}C$ to $0^{\circ}C$ is $Q_2 = m_w \cdot c_w \cdot \Delta T = 5\, kg \cdot 1\, kcal/kg/^{\circ}C \cdot 20^{\circ}C = 100\, kcal$.Step $3$: Remaining heat available for melting ice is $Q_{rem} = Q_2 - Q_1 = 100\, kcal - 20\, kcal = 80\, kcal$.Step $4$: Mass of ice that melts is $m_{melt} = Q_{rem} / L_f = 80\, kcal / 80\, kcal/kg = 1\, kg$.Step $5$: Final mass of water = Initial mass of water + Mass of melted ice = $5\, kg + 1\, kg = 6\, kg$.

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