Question

how much heat in kj is required to convert 10.4 grams of ice at -19.9 celsius to steam at 100.0 celsius? (the specific heat of ice is 2.108 j/g °c) answer with number only. no units

Explanation:

Step1: Calculate heat to raise ice temperature to 0°C

Use formula \(Q = mc\Delta T\). Here, \(m = 10.4\ g\), \(c = 2.108\ J/g^{\circ}C\), \(\Delta T=0 - (- 19.9)=19.9^{\circ}C\).
\(Q_1=10.4\times2.108\times19.9\)
\(Q_1 = 10.4\times41.9492=436.27168\ J\)

Step2: Calculate heat for phase - change (ice to water)

Latent heat of fusion of ice \(L_f = 334\ J/g\).
\(Q_2=m\times L_f=10.4\times334 = 3473.6\ J\)

Step3: Calculate heat to raise water temperature to 100°C

Specific heat of water \(c = 4.184\ J/g^{\circ}C\), \(\Delta T = 100 - 0=100^{\circ}C\)
\(Q_3=10.4\times4.184\times100=4351.36\ J\)

Step4: Calculate heat for phase - change (water to steam)

Latent heat of vaporization of water \(L_v=2260\ J/g\)
\(Q_4 = m\times L_v=10.4\times2260=23504\ J\)

Step5: Sum up all heats

\(Q_{total}=Q_1 + Q_2+Q_3+Q_4\)
\(Q_{total}=436.27168+3473.6 + 4351.36+23504\)
\(Q_{total}=31765.23168\ J\)
Convert to \(kJ\): \(Q_{total}=\frac{31765.23168}{1000}=31.76523168\approx31.8\)

Answer:

\(31.8\)