Calculate the energy in the form of heat (in kJ)required to change 71.8 g of liquid water at 25.7 $\circ$ C to ice at -16.1 $\circ$ C.Assume that no energy in the form of heat is transferred to the environment.(Heat of fusion = 333 J/g; heat of vaporization = 2256 J/g; specific heat capacities: ice = 2.06 J/g.K,liquid water = 4.184 J/g.K)

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The total energy required is the sum of energy to cool water to 0°C, freeze it, and then cool the ice to -16.1°C. Calculate each step and sum:  1. Cool water from 25.7°C to 0°C: $ q_1 = m \cdot c_{\text{water}} \cdot \Delta T = 71.8 \, \text{g} \cdot 4.184 \, \text{J/g.K} \cdot (0 - 25.7) \, \text{K} = -7716.5 \, \text{J} $.  2. Freeze water at 0°C: $ q_2 = m \cdot \text{heat of fusion} = 71.8 \, \text{g} \cdot 333 \, \text{J/g} = -23909.4 \, \text{J} $.  3. Cool ice from 0°C to -16.1°C: $ q_3 = m \cdot c_{\text{ice}} \cdot \Delta T = 71.8 \, \text{g} \cdot 2.06 \, \text{J/g.K} \cdot (-16.1) \, \text{K} = -2382.5 \, \text{J} $.  Total energy: $ q_{\text{total}} = q_1 + q_2 + q_3 = -34008.4 \, \text{J} = -34.0 \, \text{kJ} $.

Calculate the Energy in the Form of Heat (In KJ)required ∘\circ∘

Calculate the Energy in the Form of Heat (In KJ)required $\circ$