All Solutions
Page 331: Section Review
– $m = 0.050 text{kg}$
– $T_1 = 80.0 degree text{C}$
– $T_2 = 100.0 degree text{C}$
– $T_3 = 110.0 degree text{C}$
– $C_text{water} = 4180 frac{text{J}}{text{kg} cdot text{K}}$
– $C_text{steam} = 2020 frac{text{J}}{text{kg} cdot text{K}}$
– $H_v = 2.26 times 10^6 frac{text{J}}{text{kg}}$
– the water is first heated from $T_1$ to $T_2$;
– then evaporation of water at a temperature of $T_2 = 100 degree text {C}$ occurs;
– and finally the water vapor is superheated to a temperature of $T_3$.
Now we will analyze each process separately.
$$
Q_1 = mC_text{water}(T_2 – T_1)
$$
Plugging in the given values, we have:
$$begin{aligned}
Q_1 &= (0.050) cdot (4180) cdot (100.0 – 80.0)\
& = 4180 text{J}
end{aligned}$$
$$
Q_2 = mH_v
$$
Plugging in the given values, we have
$$begin{aligned}
Q_2 &= (0.050) cdot (2.26 times 10^6)\
&= 1.13 times 10^5 text{J}
end{aligned}$$
$$
Q_3 = mC_text{steam}(T_3 – T_2)
$$
Plugging in the given values, we have:
$$begin{aligned}
Q_3 &= (0.050) cdot (2020) cdot (110 – 100)\
& = 1010 text{J}
end{aligned}$$
$$begin{aligned}
Q_text{total} &= Q_1 + Q_2 + Q_3\
& = 4180 + 1.13 times 10^5 + 1010\
&=118190 text{ J}\
&= boxed{118.2 text{kJ}}
end{aligned}$$
textbf{Given:} \
$m = 1.0 text{kg}$ \
$C_text{mercury} = 140 frac{text{J}}{text{kg} cdot text{K}}$ \
$H_v = 3.06 times 10^5 frac{text{J}}{text{kg}}$ \
$T_i = 10.0 ^circ text{C}$ \
$T_f = 357 ^circ text{C}$
textbf{Calculation:}\
The heat required to heat mercury from $T_i$ to $T_f$ is given by the equation
$$
Q_1 = mC_text{mercury}(T_f – T_i)
$$
Pluggging in the given values, we have
$$
Q_1 = (1.0) cdot (140) cdot (357 – 10.0)
$$
$$
Q_1 = 48580 text{J}
$$
The heat required to vaporize mercury is given by the equation
$$
Q_2 = mH_V
$$
Plugging in the given values, we have
$$
Q_2 = (1.0) cdot (3.06 times 10^5)
$$
$$
Q_2 =3.06 times 10^5 text{J}
$$
The total heat required can now be calculated as follows
$$
Q_text{total} = Q_1 + Q_2
$$
$$
Q_text{total} = 48580 + 3.06 times 10^5
$$
$$
boxed{Q_text{total} = 3.5 times 10^5 text{J}}
$$
Q_text{total} = 3.5 times 10^5 text{J}
$$
$Q = (1.0 kg) (140 J/kg. text{textdegree} C) (357 text{textdegree} C – 10.0 text{textdegree} C) + (1.0kg)(3.06 times 10^{5} J / kg )$
$Q = 3.5 times 10^{5} J$
Mass of the hammer: $m_H=320:mathrm{kg}$
Mass of the lead block: $m_L=3: mathrm{kg}$
Velocity of the hammer: $v=5:mathrm{frac{m}{s}}$
Specific heat of lead: $c=130:mathrm{frac{J}{kg:K}}$
Change in temperature: $Delta T=5:{}^mathrm{o}C$
At the moment of impact some of the hammers kinetic energy is transferred as heat to the lead block and we will now calculate how much energy has been transferred.
$$begin{align*}
E_k&=frac{m_H v^2}{2}\
&=frac{320cdot 5^2}{2}\
&=4:mathrm{kJ}
end{align*}$$
$$begin{align*}
Delta U&=m_LcDelta T\
&=3cdot130cdot5\
&=1.95:mathrm{kJ}
end{align*}$$
$$
E_k-Delta U=2.05:mathrm{kJ}
$$
and that is why the temperature of the lead block got larger.
textbf{Given:} \
$h = 125.0 text{m}$ \
$C_text{water} = 4180 frac{text{J}}{text{kg} cdot text{K}}$ \
$g = 9.8 frac{text{m}}{text{s}^2}$
textbf{Calculation:}\
Since the potential energy of the water is all converted to thermal energy, we can express this relationship using the equation
$$
PE = Q
$$
We know that $PE = mgh$ and $Q = mC_text{water}Delta T$, thus we can express the equation above as follows
$$
mgh = mC_text{water}Delta T
$$
Isolating $Delta T$ on one side of the equation, we have
$$
Delta T = frac{gh}{C_text{water}}
$$
Plugging in the given values
$$
Delta T = frac{(9.8)(125.0)}{4180}
$$
$$
boxed{Delta T = 0.2931 ^circ text{C}}
$$
Delta T = 0.2931 ^circ text{C}
$$
$mgh = m C Delta T$ =$>$ $Delta T = dfrac{gh}{C}$
$Delta T = dfrac{(9.8 m/s^{2})(125.0m)}{4180 J /kg.text{textdegree} C}$
$Delta T = 0.293 text{textdegree} C$
Now consider the natural gas used to heat the house. During the combustion process of natural gas, heat is released and is transferred to the indoor air through the heating system. This heated indoor air does not have the ability to do the work nor can the heat received by the air be further utilized. On the other hand, the initial molecules of natural gas before combustion had a much greater potential to obtain useful work or utilize heat energy. Therefore, it is considered that heating a house with natural gas is a process in which there is a lack of usable energy, ie an increase in entropy. Consequently, the amount of disorder is increased.
