Question

1. how much heat (in kilojoules) is required to warm 10.0 g of ice, initially at $-10.0^{circ}c$ to steam at $110.0^{circ}c$? the heat capacity of ice is $2.09 j g^{-1}^{circ}c^{-1}$, and that of steam is $2.01 j g^{-1}^{circ}c^{-1}$. sketch a heating curve showing temperature versus heat added for the process described.

Explanation:

Step1: Heat required to warm ice from \(-10.0^{\circ}C\) to \(0^{\circ}C\)

Use the formula \(q = mc\Delta T\). Here, \(m = 10.0\space g\), \(c=2.09\space J\space g^{-1}\space^{\circ}C^{-1}\), \(\Delta T=0 - (- 10.0)=10.0^{\circ}C\)
\(q_1=10.0\times2.09\times10.0 = 209\space J\)

Step2: Heat required to melt ice at \(0^{\circ}C\)

The heat of fusion of water \( \Delta H_{fus}=334\space J\space g^{-1}\). Using \(q = m\Delta H_{fus}\), with \(m = 10.0\space g\)
\(q_2=10.0\times334=3340\space J\)

Step3: Heat required to warm water from \(0^{\circ}C\) to \(100^{\circ}C\)

The specific heat capacity of water \(c = 4.18\space J\space g^{-1}\space^{\circ}C^{-1}\), \(m = 10.0\space g\), \(\Delta T=100 - 0=100^{\circ}C\)
Using \(q = mc\Delta T\), \(q_3=10.0\times4.18\times100 = 4180\space J\)

Step4: Heat required to vaporize water at \(100^{\circ}C\)

The heat of vaporization of water \(\Delta H_{vap}=2260\space J\space g^{-1}\). Using \(q = m\Delta H_{vap}\), with \(m = 10.0\space g\)
\(q_4=10.0\times2260 = 22600\space J\)

Step5: Heat required to warm steam from \(100^{\circ}C\) to \(110.0^{\circ}C\)

Using \(q = mc\Delta T\), where \(m = 10.0\space g\), \(c = 2.01\space J\space g^{-1}\space^{\circ}C^{-1}\), \(\Delta T=110.0 - 100=10.0^{\circ}C\)
\(q_5=10.0\times2.01\times10.0=201\space J\)

Step6: Total heat required

\(q_{total}=q_1 + q_2+q_3 + q_4+q_5\)
\(q_{total}=209+3340 + 4180+22600+201\)
\(q_{total}=30530\space J\)
Convert to kilojoules: \(q_{total}=\frac{30530}{1000}=30.5\space kJ\)

Answer:

\(30.5\space kJ\)