Respuesta :
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.
[tex]\large {\boxed{Q = m \times c \times \Delta t} }[/tex]
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:
[tex]Q = mL_v[/tex]
[tex]9.15 \times 10^5 = m(2268)[/tex]
[tex]m = 9.15 \times 10^5 \div 2268[/tex]
[tex]m \approx 403 \texttt{ g}[/tex]
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Next, we will calculate the volume of water by using density formula as follows:
[tex]\rho = m \div V[/tex]
[tex]V = m \div \rho[/tex]
[tex]V = 403 \div 1.00[/tex]
[tex]V \approx 403 \texttt{ mL}[/tex]
[tex]V \approx 0.403 \texttt{ L}[/tex]
[tex]\texttt{ }[/tex]
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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
The volume of water vaporized by the energy from the candy bar is [tex]\boxed{403.44\,{\text{ml}}}[/tex] or [tex]\boxed{0.403\,l}[/tex].
Further Explanation:
Given:
The amount of energy given by the candy bar is [tex]915\,{\text{kJ}}[/tex].
The density of the water is [tex]1.0\,{{\text{g}} \mathord{\left/ {\vphantom {{\text{g}} {{\text{ml}}}}} \right. \kern-\nulldelimiterspace} {{\text{ml}}}}[/tex].
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.
[tex]\boxed{Q = m{L_v}}[/tex]
Here, [tex]Q[/tex] is the energy given by candy bar, [tex]m[/tex] is the mass of the water evaporated and [tex]{L_v}[/tex] is the latent heat of vaporization.
Substitute [tex]915\,{\text{kJ}}[/tex] for [tex]Q[/tex] and [tex]2268\,{{\text{J}} \mathord{\left/{\vphantom {{\text{J}} {\text{g}}}} \right.\kern-\nulldelimiterspace} {\text{g}}}[/tex] for [tex]{L_v}[/tex] in above expression.
[tex]\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}[/tex]
Convert the mass of the water into the volume.
[tex]V = \dfrac{m}{\rho }[/tex]
Substitute [tex]403.44\,{\text{g}}[/tex] for [tex]m[/tex] and [tex]1.0\,{{\text{g}} \mathord{\left/{\vphantom {{\text{g}} {{\text{ml}}}}} \right.\kern-\nulldelimiterspace} {{\text{ml}}}}[/tex] for [tex]\rho[/tex] in above expression.
[tex]\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}[/tex]
Thus, the volume of water vaporized by the energy from the candy bar is [tex]\boxed{403.44\,{\text{ml}}}[/tex] or [tex]\boxed{0.403\,l}[/tex].
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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.
