Answer:
[tex]\text{36480cal}[/tex]
Explanation:
[tex]\text{Solution:}\\\text{Given:}\\\text{Mass of water (m)}=60\text{g}\\\text{Initial temperature(t}_1)=32^\circ\text{C}\\\text{Final temperature(t}_2)=100^\circ\text{C}\\\text{Latent heat of vaporization (L}_\text{f})=540\text{cal/g.}\\\text{We know,}\\\text{Specific heat capacity of water(S)}=1\text{cal/g}[/tex]
[tex]\text{Now,}\\\text{Heat required for water to reach 100}^\circ\text{C}(\text{H}_1)=\text{ms}\Delta\text{T}=60\times1\times(100-32)\\\text{or, H}_1=4080\text{cal}[/tex]
[tex]\text{Heat required to convert water to steam(H}_2)=\text{mL}_\text{f}=60\times540=32400\text{cal}[/tex]
[tex]\therefore\ \text{Total heat required = H}_1\text{+ H}_2=4080+32400=36480\text{cal}[/tex]
To completely convert 60 grams of water at 100 °C to steam, 32,400 calories of heat are required, calculated using the product of the mass of water and the latent heat of vaporization.
The question asks how much heat is required to completely convert 60 grams of water at 100 °C to steam, given the latent heat of vaporization of water is 540 cal/g. To find the answer, we use the formula for calculating the heat absorbed during the phase change from liquid to gas without a change in temperature, which is:
Heat (Q) = mass (m) × latent heat of vaporization (L)
Substituting the given values:
Q = 60g × 540 cal/g = 32,400 calories
Therefore, the heat required to vaporize 60 grams of water at 100°C into steam is 32,400 calories.
