# How much heat (in kJ) is required to warm 10.0 g of ice, initially at -15.0 C, to steam at 111.0...

## Question:

How much heat (in kJ) is required to warm 10.0 g of ice, initially at -15.0 {eq}^{\circ} {/eq}C, to steam at 111.0 {eq}^{\circ} {/eq}C? The heat capacity of ice is 2.09 J/g{eq}\cdot ^{\circ} {/eq}C and that of steam is 2.01 J/g{eq}\cdot ^{\circ} {/eq}C.

## Heating Curve

The slopped region in a heating curve is when there is only one phase present and the energy added is used to increases the temperature. At the plateau region, two phases exits and the energy added is used in phase change at a constant temperature.

## Answer and Explanation:

**30.7 kJ of heat are required.**

Several steps are involved:

1. heat ice from -15 {eq}^oC {/eq} to 0 {eq}^oC {/eq} (use heat capacity of ice, 2.09 J/g{eq}\cdot ^{\circ} {/eq}C)

2. melt the ice (use enthalpy of fusion of water, 334 J/g)

3. heat water from 0 {eq}^oC {/eq} to 100 {eq}^oC {/eq} (use heat capacity of water, 4.18 J/g{eq}\cdot ^{\circ} {/eq}C)

4. boiling water (use enthalpy of vaporization of water, 2260 J/g)

5. heat steam from 100 {eq}^oC {/eq} to 111.0 {eq}^oC {/eq} (use heat capacity of steam, 2.01 J/g{eq}\cdot ^{\circ} {/eq}C)

Here are the calculation of each step

1. $$q = mc \Delta T\\ q = (10.0 \ g)(2.09 \dfrac {J}{g \cdot ^oC } )(0 - -15.0 ^oC) = 313.5 \ J $$

2. $$q = m \Delta H_{fus}\\ q = (10.0 \ g)(334 \ J/g) = 3340 \ J $$

3. $$q = mc \Delta T\\ q = (10.0 \ g)(4.18 \dfrac {J}{g \cdot ^oC } )(100 - 0 ^oC) = 4180 \ J $$

4. $$q = m \Delta H_{fus}\\ q = (10.0 \ g)(2260 \ J/g) = 22600 \ J $$

5. $$q = mc \Delta T\\ q = (10.0 \ g)(2.01 \dfrac {J}{g \cdot ^oC } )(111.0 - 100 ^oC) = 221.1 \ J $$

Adding all the energy involved in each step gives

$$313.5 \ J + 3340 \ J + 4180 \ J + 22600 \ J + 221.1 \ J =30654 J = \boxed {30.7 \ kJ} $$

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