Answer :
The final temperature of the water is about 21°C
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!
Given:
initial temperature of water = t_o = 20°C
specific heat capacity of water = c = 4186 J/kg°C
density of water = ρ = 1000 kg/m³
volume of water = V = 5 dm³ = 5 × 10⁻³ m³
mass of person = m₁ = 90 kg
height of tower = h = 30 m
gravitational acceleration = g = 9.8 m/s²
Asked:
final temperature of the water = t = ?
Solution:
Firstly, I have corrected the volume of water to 5 dm³ to make it more reasonable.
We can calculate the final temperature of the water by using Conservation of Energy as shown below:
[tex]\texttt{Potential Energy of The Person = Heat Energy of Water}[/tex]
[tex]E_p = Q[/tex]
[tex]m_1gh = m_2c \Delta t[/tex]
[tex]m_1gh = \rho V c \Delta t[/tex]
[tex]m_1gh = \rho V c (t - t_o)[/tex]
[tex](m_1gh) \div (\rho V c) = (t - t_o)[/tex]
[tex]t = t_o + [ (m_1gh) \div (\rho V c) ][/tex]
[tex]t = 20 + [ (90(9.8)(30)) \div (1000 ( 5 \times 10^{-3} ) ( 4186) ) ][/tex]
[tex]t = 20 + [ 26460 \div 20930 ][/tex]
[tex]t \approx 21^o C[/tex]
[tex]\texttt{ }[/tex]
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Answer details
Grade: College
Subject: Physics
Chapter: Thermal Physics
Keywords: Heat , Temperature , Block , Aluminium , Ice , Cold , Water
The final temperature of the water will be equal to 21°C
What is specific heat ?
specific heat, the quantity of heat required to raise the temperature of one gram of a substance by one Celsius degree.
Specific Heat Capacity is the amount of energy needed to raise temperature of 1 kg body for 1°C.
[tex]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 )
It is given that
initial temperature of water = t_o = 20°C
specific heat capacity of water = c = 4186 J/kg°C
density of water = ρ = 1000 kg/m³
volume of water = V = 5 dm³ = 5 × 10⁻³ m³
mass of person = m₁ = 90 kg
height of tower = h = 30 m
gravitational acceleration = g = 9.8 m/s²
The final temeperature of the water will be equal to:
Firstly, I have corrected the volume of water to 5 dm³ to make it more reasonable.
We can calculate the final temperature of the water by using Conservation of Energy as shown below:
Potential energy of a person = Heat energy of water
[tex]m_1gh=m_2c\Delta t[/tex]
[tex]m_1gh=\rho Vc\Delta t[/tex]
[tex]m_1gh=\rhoVc(t-t_o)[/tex]
[tex](t-t_o)=\dfrac{m_1gh}{\rho Vc}[/tex]
[tex]t=t_o+\dfrac{m_1gh}{\rho Vc}[/tex]
[tex]t=t_o+\dfrac{(90(9.8)(30)}{(1000(5\times10^{-3})(4186)}[/tex]
[tex]t=20+\dfrac{26460}{20930}[/tex]
[tex]t=20^oc[/tex]
Thus the final temperature of the water will be equal to 21°C
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