m g of water at 30^{circ} C is mixed with 5 g | vedclass

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Question

(B) According to the principle of calorimetry,Heat lost by water = Heat gained by ice.Let $m$ be the mass of water in grams.Heat lost by water $= m

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) According to the principle of calorimetry,Heat lost by water = Heat gained by ice.Let $m$ be the mass of water in grams.Heat lost by water $= m 	imes c_w 	imes (T_i - T_f) = m 	imes 1 	imes (30 - 6) = 24m 	ext{ cal}$.Heat gained by ice consists of three parts:$1$. Heating ice from $-20^{\circ} C$ to $0^{\circ} C$: $Q_1 = m_i 	imes c_i 	imes \Delta T = 5 	imes 0.5 	imes (0 - (-20)) = 5 	imes 0.5 	imes 20 = 50 	ext{ cal}$.$2$. Melting ice at $0^{\circ} C$: $Q_2 = m_i 	imes L_f = 5 	imes 80 = 400 	ext{ cal}$.$3$. Heating the melted water from $0^{\circ} C$ to $6^{\circ} C$: $Q_3 = m_i 	imes c_w 	imes \Delta T = 5 	imes 1 	imes (6 - 0) = 30 	ext{ cal}$.Total heat gained $= 50 + 400 + 30 = 480 	ext{ cal}$.Equating heat lost and gained: $24m = 480$.$m = 480 / 24 = 20 	ext{ g}$.

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