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Q.When 0.15 kg of ice at 0 degrees C is mixed with 0.30 kg of water at 50 degrees C in a container, the resulting temperature is 6.7 degrees C. Calculate the heat of fusion of ice. (S_water = 4186 J kg^-1 K^-1) OR A pan filled with hot food cools from 94 degrees C to 86 degrees C in 2 minutes when the room temperature is 20 degrees C. How long will it take to cool from 71 degrees C to 69 degrees C?

Madhya Pradesh MpbseMP Board Higher Secondary (Class 11) 2020Subjective· 4mImportance★★★★★
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Equating heat lost by the water to heat gained by the ice (melting + warming) and solving for L gives about 3.34 x 10^5 J/kg, matching the standard textbook value.

Given: mass of ice m_ice = 0.15 kg at 0 degC, mass of water m_w = 0.30 kg at 50 degC, final (mixture) temperature = 6.7 degC, specific heat of water s_w = 4186 J kg^-1 K^-1.

Heat lost by the water as it cools from 50 degC to 6.7 degC:

Q_lost = m_w s_w (50 - 6.7) = 0.30 x 4186 x 43.3 = 54,376 J (approx)

Heat gained by the ice has two parts: (i) melting at 0 degC, requiring m_ice L, and (ii) the melted ice-water warming from 0 degC to 6.7 degC:

Q_gained = m_ice L + m_ice s_w (6.7 - 0) = 0.15 L + (0.15 x 4186 x 6.7) = 0.15 L + 4,207 J (approx)

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