Question

How long would it take for 1.50 mol of water at 100.0 ∘c to be converted completely into steam if heat were added at a constant rate of 24.0 j/s ? express your answer to three significant figures and include the appropriate units?

Answer 1

The total heat that is needed by the water to get converted to steam is calculated below through the equation,

     H = (Hv)(n)

where H is the heat, Hv is the heat of vaporization (40.67 kJ/mol), and n is the number of mols. Substituting the known values,

   H = (40.66 kJ/mol)(1.5 mol) = 60.99 kJ

Calculating for the time,

    t = (60990 J) / (24 J/s)
    t = 2541.25 s

Answer: 2541.25 s or approximately 42.35 minutes

Answer 2

The time taken to completely convert the water into the steam is $\boxed{2541\,{\text{s}}}$ or $\boxed{42.4\,\min }$.

Further Explanation:

The heat required to completely convert the water at $100\,{\text{^\circ C}}$ into the steam is defined as the latent heat of vaporization. The latent heat of vaporization of the water is $40.66\,{{{\text{kJ}}} \mathord{\left/{\vphantom {{{\text{kJ}}} {{\text{mol}}}}} \right.\kern-\nulldelimiterspace} {{\text{mol}}}}$.

The expression for the amount of heat required for conversion of $1.5\,{\text{mol}}$ water into steam is.

$Q = n \times L$  

Substitute $1.5\,{\text{mol}}$ for $n$ and $40.66\,{{{\text{kJ}}} \mathord{\left/{\vphantom {{{\text{kJ}}} {{\text{mol}}}}} \right.\kern-\nulldelimiterspace} {{\text{mol}}}}$ for$L$ in above expression.

$\begin{aligned}Q&= 1.5\,{\text{mol}}\times 40.66\,{{{\text{kJ}}} \mathord{\left/{\vphantom {{{\text{kJ}}} {{\text{mol}}}}} \right.\kern-\nulldelimiterspace} {{\text{mol}}}} \\&= 60.99\,{\text{kJ}}\\\end{aligned}$  

The rate of addition of heat to the water is $24\,{{\text{J}} \mathord{\left/{\vphantom {{\text{J}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$.

Therefore, the time taken in conversion of the water into the steam is:

 $t = \dfrac{Q}{r}$

Here, $r$ is the rate of heat addition to the water.

Substitute the values of $Q$ and $r$.

$\begin{aligned}t&= \frac{{60.99 \times 1000\,{\text{J}}}}{{24\,{{\text{J}} \mathord{\left/{\vphantom {{\text{J}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}} \\&= 2541.25\,{\text{s}}\left( {\frac{{1\,\min }}{{60\,{\text{s}}}}} \right) \\&\approx 42.4\,\min\\\end{aligned}$  

Thus, the time taken to completely convert the water into steam is $\boxed{2541\,{\text{s}}}$ or $\boxed{42.4\,\min }$.

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Answer Details:

Grade: College

Chapter: Heat transfer

Subject: Physics

Keywords: 1.50 mol, water, 100 C, constant rate, converted completely, steam, latent heat, vaporization, heat added, time taken, 42.25 min.