All Solutions
Page 325: Practice Problems
textbf{Given:} \
$m = 0.100 text{kg}$ \
$T_i = -20.0 ^circ text{C}$ \
$T_f = 0.0 ^circ text{C}$ \
$H_f = 3.34 times 10^5 frac{text{J}}{text{kg}}$ \
$C_text{water} = 2060 frac{text{J}}{text{kg} cdot text{K}}$ \
textbf{Calculation:}\
We first calculate the heat required to raise the temperature of water from $-20.0 ^circ text{C}$ to $0.0 ^circ text{C}$ using the equation
$$
Q_1 = mC_text{water}(T_f – T_i)
$$
Plugging in the given values, we have
$$
Q_1 = (0.100) cdot (2060) cdot (0.00 – (-20.0))
$$
$$
Q_1 = 4120 text{J}
$$
We now calculate the heat required to melt the ice using the equation below
$$
Q_2 = mH_f
$$
Plugging in the given $m$ and $H_f$, we have
$$
Q_2 = (0.100) cdot (3.34 times 10^5)
$$
$$
Q_2 = 33400 text{J}
$$
To get the total amount of heat needed, we just need to add the calculated $Q_1$ and $Q_2$
$$
Q_text{total} = Q_1 + Q_2
$$
$$
Q_text{total} = 4120 + 33400
$$
$$
boxed{Q_text{total} = 3.75 times 10^4 text{J}}
$$
Q_text{total} = 3.75 times 10^4 text{J}
$$
$Q_{heat ice} = m c_{ice} Delta T$
$Q_{heat ice} = (0.100 Kg)*(2060 J/Kg.C)*(0 – (-20))$
$Q_{heat ice} = 4120 J$
Calculate the heat needed to melt ice:
$Q_{melt} = m H_f$
$Q_{melt} = (0.100)*(3.34e5)$
$Q_{melt} = 33400 J$
The total amount of heat is:
$$
Q = Q_{heat ice} + Q_{melt} = 4120 + 33400 = 3.75 times 10^4 J
$$
3.75 times 10^4 J
$$
textbf{Given:} \
$m = 0.200 text{kg}$ \
$T_1 = 60.0 ^circ text{C}$ \
$T_2 = 100.0 ^circ text{C}$ \
$T_3 = 140.0 ^circ text{C}$ \
$H_v = 2.26 times 10^6 frac{text{J}}{text{kg}}$ \
$C_text{water} = 4180 frac{text{J}}{text{kg} cdot text{K}}$ \
$C_text{steam} = 2020 frac{text{J}}{text{kg} cdot text{K}}$ \
textbf{Calculation:} \
We first calculate the heat required to raise the temperature of water from $60.0 ^circ text{C}$ to $100.0 ^circ text{C}$ using the equation
$$
Q_1 = mC_text{water}(T_2 – T_1)
$$
Pluggging in the given values, we have
$$
Q_1 = (0.200) cdot (4180) cdot (100.0 – 60.0)
$$
$$
Q_1 = 33440 text{J}
$$
We now calculate the heat required to vaporize water into steam using the equation below
$$
Q_2 = mH_v
$$
Pluggging in the given $m$ and $H_f$, we have
$$
Q_2 = (0.200) cdot (2.26 times 10^6)
$$
$$
Q_2 = 452000 text{J}
$$
We now calculate the heat required to raise the temperature of water from $100.0 ^circ text{C}$ to $140.0 ^circ text{C}$ using the equation
$$
Q_3 = mC_text{steam}(T_3 – T_2)
$$
Pluggging in the given values, we have
$$
Q_3 = (0.200) cdot (2020) cdot (140.0 – 100.0)
$$
$$
Q_3 = 16160 text{J}
$$
To get the total amount of heat needed, we just need to add the calculated $Q_1$, $Q_2$, and $Q_3$
$$
Q_text{total} = Q_1 + Q_2 + Q_3
$$
$$
Q_text{total} = 33440 + 452000 + 16160
$$
$$
boxed{Q_text{total} = 5.02 times 10^5 text{J}}
$$
Q_text{total} = 5.02 times 10^5 text{J}
$$
$Q_{heat water} = m c_{water} Delta T$
$Q_{heat water} = (0.200 Kg)*(4180 J/Kg.C)*(100 – 60)$
$Q_{heat water} = 33440 J$
Calculate the heat needed to vaporize water:
$Q_{vaporization} = m H_v$
$Q_{vaporization} = (0.200)*(2.26e6)$
$Q_{vaporization} = 452000 J$
Calculate the heat needed to raise the vapor temperature:
$Q_{heat vapor} = m c_{vapor} Delta T$
$Q_{heat vapor} = (0.200 Kg)*(2020 J/Kg.C)*(140 – 100)$
$Q_{heat vapor} = 16160 J$
The total amount of heat is:
$$
Q = 33440 + 452000 + 16160 = 5.02 times 10^5 J
$$
5.02 times 10^5 J
$$
textbf{Given:} \
$m = 0.3 text{kg}$ \
$T_1 = -30.0 ^circ text{C}$ \
$T_2 = 0.0 ^circ text{C}$ \
$T_2 = 100.0 ^circ text{C}$ \
$T_4 = 130.0 ^circ text{C}$ \
$H_f = 3.34 times 10^5 frac{text{J}}{text{kg}}$ \
$H_v = 2.26 times 10^6 frac{text{J}}{text{kg}}$ \
$C_text{ice} = 2060 frac{text{J}}{text{kg} cdot text{K}}$ \
$C_text{water} = 4180 frac{text{J}}{text{kg} cdot text{K}}$ \
$C_text{steam} = 2020 frac{text{J}}{text{kg} cdot text{K}}$ \
textbf{Calculation:}\
We first calculate the heat required to increase the temperature of ice from $T_1$ to $T_2$ using the equation below
$$
Q_1 = mC_text{ice}(T_2 – T_1)
$$
Plugging in the given values, we have
$$
Q_1 = (0.3) cdot (2060) cdot (0.0 – (-30.0))
$$
$$
Q_1 = 18540 text{J}
$$
We calculate the heat required to transform ice to liquid water using the equation below
$$
Q_2 = mH_f
$$
Plugging in the given values, we have
$$
Q_2 = (0.3) cdot (3.34 times 10^5)
$$
$$
Q_2 = 100200 text{J}
$$
We calculate the heat required to increase the temperature of liquid water from $T_2$ to $T_3$ using the equation below
$$
Q_3 = mC_text{water}(T_3 – T_2)
$$
Plugging in the given values, we have
$$
Q_3 = (0.3) cdot (4180) cdot (100.0 – 0.0)
$$
$$
Q_3 = 125400 text{J}
$$
We calculate the heat required to transform liquid water to steam using the equation below
$$
Q_4 = mH_v
$$
Plugging in the given values, we have
$$
Q_4 = (0.3) cdot (2.26 times 10^6)
$$
$$
Q_4 = 678000 text{J}
$$
We calculate the heat required to increase the temperature of steam from $T_3$ to $T_4$ using the equation below
$$
Q_5 = mC_text{steam}(T_4 – T_3)
$$
Plugging in the given values, we have
$$
Q_5 = (0.3) cdot (2020) cdot (130.0 – 100.0)
$$
$$
Q_5 = 18180 text{J}
$$
Finally, we determine the total heat by adding all the calculated heat from $Q_1$ to $Q_5$
$$
Q_text{total} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5
$$
$$
Q_text{total} = 18540 + 100200 + 125400 + 678000 + 18180
$$
$$
boxed{Q_text{total} = 9.40 times 10^5 text{J}}
$$
Q_text{total} = 9.40 times 10^5 text{J}
$$
$Q_{heat ice} = m c_{ice} Delta T$
$Q_{heat ice} = (0.300 Kg)*(2060 J/Kg.C)*((0.00) – (-30))$
$Q_{heat ice} = 18540 J$
Calculate the heat needed to melt ice:
$Q_{melt} = m H_f$
$Q_{melt} = (0.300)*(3.34e5)$
$Q_{melt} = 100200 J$
$Q_{heat water} = m c_{water} Delta T$
$Q_{heat water} = (0.300 Kg)*(4180 J/Kg.C)*(100 – 0.00)$
$Q_{heat water} = 125400 J$
Calculate the heat needed to vaporize water:
$Q_{vaporization} = m H_v$
$Q_{vaporization} = (0.300)*(2.26e6)$
$Q_{vaporization} = 678000 J$
Calculate the heat needed to raise the vapor temperature:
$Q_{heat vapor} = m c_{vapor} Delta T$
$Q_{heat vapor} = (0.300 Kg)*(2020 J/Kg.C)*(130 – 100)$
$Q_{heat vapor} = 18180 J$
$$
Q = 18540 + 100200 + 125400 + 678000 + 18180 = 9.40 times 10^5 J
$$
9.40 times 10^5 J
$$
textbf{Given:} \
$Q = 75 text{J}$ \
textbf{Calculation:}\
We know that the 1st Law of Thermodynamics is given by the equation below
begin{equation}
Delta U = Q – W
end{equation}
Since the balloon stays at the same temperature after expansion, its internal energy must not change
$$
Delta U = 0
$$
Plugging this into Equation (1), we have
$$
0 = Q – W
$$
Isolating $W$ on one side of the equation
$$
W = Q
$$
Plugging in the given $Q$
$$
boxed{W = 75 text{J}}
$$
W = 75 text{J}
$$
$Delta U = Q – W$
The change in thermal energy of an object is equal to the heat added to the
object minus the work done by the object.
The temperature of the balloon is the same, therefore its thermal energy is the same: $Delta U = 0$.
$Delta U = 0 = Q – W$
The balloon absorbed 75 J of heat: $Q=75 J$:
$0 = Q – W$
$0 = 75 – W$
Thus:
$$
W = 75 J
$$
75 J
$$
textbf{Given:} \
$m = 0.40 text{kg}$ \
$Delta T = 5.0 ^circ text{C}$ \
$C_text{aluminum} = 897 frac{text{J}}{text{kg} cdot text{K}}$ \
textbf{Calculation:}\
Since the work done by the drill is equal to the heat added to the aluminum, we can write it as follows
$$
W = Q = mC_text{aluminum}Delta T
$$
Plugging in the given values, we have
$$
W = (0.40) cdot (897) cdot (5.0)
$$
$$
boxed{W = 1800 text{J}}
$$
W = 1800 text{J}
$$
$W = (0.40 Kg) (897 J/kg.C) (5.0 C)$
$$
W = 1800 J
$$
1800 J
$$
textbf{Given:} \
$m = 0.50 text{kg}$ \
$h = 1.5 text{m}$ \
$Delta T = 1.0 ^circ text{C}$ \
$g = 9.8 frac{text{m}}{text{s}^2}$ \
$C_text{lead} = 130 frac{text{J}}{text{kg} cdot text{K}}$
textbf{Calculation:}\
The heat needed to increase the temperature of the lead shot by $1.0 ^circ text{C}$ is given by the equation below
$$
Q = mC_text{lead}Delta T
$$
The work done by dropping the lead shot $n$ times can be calculated as follows
$$
W_n = nW = nmgh
$$
Since $Q$ must be equal to $W_n$, we have
$$
mC_text{lead}Delta T = nmgh
$$
Isolating $n$ on one side of the equation
$$
n = frac{C_text{lead}Delta T}{gh}
$$
Plugging in the given values, we have
$$
n = frac{(130) cdot (1.0)}{(9.8) cdot (1.5)}
$$
$$
boxed{n = 9 text{drops}}
$$
n = 9 text{drops}
$$
$N m g h = m c_{lead} Delta T$
Solve for N:
$N = dfrac{c_{lead} Delta T}{g h} = dfrac{(130)*(1.0)}{(9.80)*(1.5)}$
$$
N = 8.8 drops
$$
8.8 drops
$$
textbf{Given:} \
$m = 0.15 text{kg}$ \
$Delta T = 2.0 ^circ text{C}$ \
$W = 0.050 text{J}$ \
$C_text{water} = 4180 frac{text{J}}{text{kg} cdot text{K}}$
textbf{Calculation:}\
The heat needed to raise the temperature of the lead shot is given by the equation below
$$
Q = mC_text{water}Delta T
$$
The total work done by stirring the cup of tea by $n$ times can be calculated as follows
$$
W_n = nW
$$
Since $Q$ must be equal to $W_n$, we have
$$
mC_text{water}Delta T = nW
$$
Isolating $n$ on one side of the equation
$$
n = frac{mC_text{water}Delta T}{W}
$$
Plugging in the given values, we have
$$
n = frac{(0.15) cdot (4180) cdot (2.0)}{(0.050)}
$$
$$
boxed{n = 2.6 times 10^4 text{times}}
$$
n = 2.6 times 10^4 text{times}
$$
$Delta U =(0.15 kg)(4180J/kg.text{textdegree} C)(2.0 text{textdegree} C)$
$Delta U = 1.3 times 10^{3} J$
For the number of stirs ;
$dfrac{1.3 times 10^{3} J }{0.050 J}=2.6 times 10^{4}$
$$
Delta U=Q-W
$$
Since the internal energy of the system is proportional to the temperature of the system if we want to cool down the system we need to have less internal energy at the end of the process or in other words we need to have negative change in the internal energy $Delta U<0$.
$$ W=0,:Q0$$
The first case corresponds to the system which is not doing any work nor the work is done on it (volume does not change) but it transfers heat to some colder object that it is in contact with.
The second case corresponds to the system which does not exchange heat with its surroundings but it does some work against the forces acting on it (expands).
$$
Q<W
$$
where the heat is added to the system and the work is done by the system.
