# How much energy is needed to convert 23.0 grams of ice at -10.0°C into steam at 109°C?

Converting 23.0 g of ice at -10.0 °C to steam at 109 °C requires 69.8 kJ of energy.

There are five heats to consider:

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To calculate the energy required, use the formula:

[ Q = m \cdot (c_{\text{ice}} \cdot \Delta T_{\text{ice}} + L_{\text{fusion}} + c_{\text{water}} \cdot \Delta T_{\text{melt}} + L_{\text{vaporization}} + c_{\text{steam}} \cdot \Delta T_{\text{steam}}) ]

[ Q = 23.0 , \text{g} \cdot (2.09 , \text{J/g°C} \cdot (0 - (-10.0°C)) + 334 , \text{J/g} + 4.18 , \text{J/g°C} \cdot (100.0°C - 0) + 2260 , \text{J/g} + 2.01 , \text{J/g°C} \cdot (109.0°C - 100.0°C)) ]

[ Q \approx 23.0 , \text{g} \cdot 748.79 , \text{J/g} ]

[ Q \approx 17,213.17 , \text{J} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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