Question

how much heat in kj is required to convert 23.3 grams of ice at -27 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 warm ice from -27°C to 0°C

Use formula \(Q = mc\Delta T\). Here, \(m = 23.3\space g\), \(c=2.108\space J/g^{\circ}C\), \(\Delta T=0 - (- 27)=27^{\circ}C\).
\(Q_1=23.3\times2.108\times27\)
\(Q_1 = 23.3\times56.916=1326.1428\space J\)

Step2: Calculate heat to melt ice at 0°C

Use formula \(Q = mL_f\). Latent heat of fusion \(L_f = 334\space J/g\)
\(Q_2=23.3\times334 = 7782.2\space J\)

Step3: Calculate heat to warm water from 0°C to 100°C

Specific heat of water \(c = 4.184\space J/g^{\circ}C\), \(\Delta T=100 - 0=100^{\circ}C\)
\(Q_3=23.3\times4.184\times100\)
\(Q_3=23.3\times418.4 = 9748.72\space J\)

Step4: Calculate total heat

\(Q_{total}=Q_1 + Q_2+Q_3\)
\(Q_{total}=1326.1428+7782.2 + 9748.72\)
\(Q_{total}=18857.0628\space J\)
Convert to kJ: \(Q_{total}=\frac{18857.0628}{1000}=18.8570628\approx18.9\)

Answer:

18.9