Question

The human body obtains 915 kj of energy from a candy bar. if this energy were used to vaporize water at 100.0 °c, how much water (in liters) could be vaporized? (assume the density of water is 1.00 g>ml.

Answer 1

0.403 liters of water could be vaporized

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Further explanation

Specific Heat Capacity is the amount of energy needed to raise temperature of 1 kg body for 1°C.

$\large {\boxed{Q = m \times c \times \Delta t} }$

Q = Energy ( Joule )

m = Mass ( kg )

c = Specific Heat Capacity ( J / kg°C )

Δt = Change In Temperature ( °C )

Let us now tackle the problem!

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Given:

heat added = Q = 915 kJ = 9.15 × 10⁵ J

specific latent heat of vaporization of water = Lv = 2268 J/g

density of water = 1.00 g/mL

Asked:

volume of water = V = ?

Solution:

We will use latent heat of vaporization formula to solve this problem as follows:

$Q = mL_v$

$9.15 \times 10^5 = m(2268)$

$m = 9.15 \times 10^5 \div 2268$

$m \approx 403 \texttt{ g}$

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Next, we will calculate the volume of water by using density formula as follows:

$\rho = m \div V$

$V = m \div \rho$

$V = 403 \div 1.00$

$V \approx 403 \texttt{ mL}$

$V \approx 0.403 \texttt{ L}$

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Answer details

Grade: College

Subject: Physics

Chapter: Thermal Physics

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Keywords: Heat , Temperature , Block , Aluminium , Ice , Cold , Water

Answer 2

The volume of water vaporized by the energy from the candy bar is $\boxed{403.44\,{\text{ml}}}$ or $\boxed{0.403\,l}$.

Further Explanation:

Given:

The amount of energy given by the candy bar is $915\,{\text{kJ}}$.

The density of the water is $1.0\,{{\text{g}} \mathord{\left/ {\vphantom {{\text{g}} {{\text{ml}}}}} \right. \kern-\nulldelimiterspace} {{\text{ml}}}}$.

Concept:

The amount of energy provided by the candy bar changes the state of the water and vaporizes it. The energy required for the evaporation is given by the formula.

 $\boxed{Q = m{L_v}}$

Here, $Q$ is the energy given by candy bar, $m$ is the mass of the water evaporated and ${L_v}$ is the latent heat of vaporization.

Substitute $915\,{\text{kJ}}$ for $Q$ and $2268\,{{\text{J}} \mathord{\left/{\vphantom {{\text{J}} {\text{g}}}} \right.\kern-\nulldelimiterspace} {\text{g}}}$ for ${L_v}$ in above expression.

$\begin{aligned}9.15 \times {10^5} &= m \times 2268 \\m&= \frac{{9.15 \times {{10}^5}}}{{2268}}\,{\text{g}}\\&= {\text{403}}{\text{.44}}\,{\text{g}} \\\end{aligned}$  

Convert the mass of the water into the volume.

$V = \dfrac{m}{\rho }$  

Substitute $403.44\,{\text{g}}$ for $m$ and $1.0\,{{\text{g}} \mathord{\left/{\vphantom {{\text{g}} {{\text{ml}}}}} \right.\kern-\nulldelimiterspace} {{\text{ml}}}}$ for $\rho$ in above expression.

$\begin{aligned}V &= \frac{{403.44\,{\text{g}}}}{{1.0\,{{\text{g}} \mathord{\left/{\vphantom {{\text{g}} {{\text{ml}}}}} \right. \kern-\nulldelimiterspace} {{\text{ml}}}}}} \\&= 403.44\,{\text{ml}}\\\end{aligned}$  

Thus, the volume of water vaporized by the energy from the candy bar is $\boxed{403.44\,{\text{ml}}}$ or $\boxed{0.403\,l}$.

Learn More:

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2. One consequence of the third law of thermodynamics is that brainly.com/question/3564634
3. John and caroline go out for a walk one day. This graph represents their distance from home. Which statement accurately describes their walk brainly.com/question/11313502

Answer Details:

Grade: High School

Chapter: Heat Transfer

Subject: Physics

Keywords:  Human body, energy, from a candy bar, latent heat of vaporization, volume of water, vaporized, evaporation, changes the state.