Question

What minimum heat is needed to bring 200 g of water at 30 ∘C to the boiling point and completely boil it away? The specific heat of water is 4190 J/(kg⋅K) and its heat of vaporization is 22.6××1055 J/kg.

Express your answer with the appropriate units.

Answer 1

Answer 2

 Calculation Steps:

1. Heat to raise the temperature from 30 °C to 100 °C (boiling point):

   Formula:

   $$Q_1=m\cdot c\cdot \mathrm{\Delta }T$$

   Calculation:

   $$Q_1=0.2\text{kg}\times 4190$$

• Mass of water, $m=200\text{g}=0.2\text{kg}$

• Specific heat of water,

• Temperature change, $\mathrm{\Delta }T=100\text{°C}-30\text{°C}=70\text{K}$

2. Heat to vaporize the water at 100 °C:

   Formula:

   $$Q_2=m\cdot L_v$$

   Calculation:

   $$Q_2=0.2\text{kg}\times 22.6\times 10^5\text{J/kg}=452,000\text{J}$$

• Heat of vaporization, $L_v=22.6\times 10^5\text{J/kg}$

3. Total heat required:

   $$Q_{\text{total}}=Q_1+Q_2=58,660\text{J}+452,000\text{J}=515,420\text{J}$$

Thus, the minimum heat needed is 515,420 J