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How much heat energy in kilojoules is required to convert 41.0g of ice at -18.0 degrees C to water at 25.0 degrees C?
Wiki User
∙ 11y ago
19.7 kJ
Wiki User
∙ 11y ago
The total heat energy required can be broken down into two steps: first, raising the temperature of ice from -18.0°C to 0°C, then melting the ice at 0°C to water at 25.0°C. The heat energy required is calculated using the formula: Q = mcΔT, where Q is heat energy, m is mass, c is specific heat capacity, and ΔT is the temperature change.
Step 1: Raise temperature of ice to 0°C Q1 = 41.0g * 2.09 J/g°C * 18.0°C = 1552.22 J
Step 2: Melt ice at 0°C Q2 = 41.0g * 334 J/g = 13694 J
Total heat energy required = Q1 + Q2 = 13694 J + 1552.22 J = 15246.22 J
Convert J to kJ: 15246.22 J * (1 kJ / 1000 J) ≈ 15.25 kJ
Therefore, approximately 15.25 kJ of heat energy is required to convert 41.0g of ice at -18.0 degrees C to water at 25.0 degrees C.
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