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(A) According to the principle of calorimetry, Heat lost by water = Heat gained by ice.Let $m_i$ be the mass of ice added.Heat lost by water: $Q_1 = m

Vedclass · Accepted Answer

$47$

Vedclass · Answer

(A) According to the principle of calorimetry, Heat lost by water = Heat gained by ice.Let $m_i$ be the mass of ice added.Heat lost by water: $Q_1 = m_w c_w \Delta T_w = (0.2 \,kg) 	imes (4190 \,J/kg^{\circ} C) 	imes (20^{\circ} C - 0^{\circ} C) = 16760 \,J$.Heat gained by ice to reach $0^{\circ} C$: $Q_2 = m_i c_i \Delta T_i = m_i 	imes (2100 \,J/kg^{\circ} C) 	imes (0^{\circ} C - (-10^{\circ} C)) = 21000 m_i \,J$.Heat gained by ice to melt at $0^{\circ} C$: $Q_3 = m_i L_F = m_i 	imes (3.34 	imes 10^5 \,J/kg) = 334000 m_i \,J$.Equating heat lost and gained: $16760 = 21000 m_i + 334000 m_i$.$16760 = 355000 m_i$.$m_i = \frac{16760}{355000} \approx 0.0472 \,kg = 47.2 \,g$.Rounding to the nearest integer, the mass of ice required is $47 \,g$.

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