All Solutions
Page 339: Standardized Test Practice
To convert $^circ text{C}$ to $text{K}$, we use the equation
begin{equation}
T_text{K} = T_text{C} + 273
end{equation}
For Option A, we plug in $T_text{C} = -273 ^circ text{C}$ to Equation (1)
$$
T_text{K} = (-273) + 273
$$
$$
T_text{K} = 0 text{K}
$$
Thus, $text{A}. -273 ^circ text{C} = 0 text{K}$ is correct so it is not the answer for this item.
For Option B, we plug in $T_text{C} = 273 ^circ text{C}$ to Equation (1)
$$
T_text{K} = 273 + 273
$$
$$
T_text{K} = 546 text{K}
$$
Thus, $text{B}. 273 ^circ text{C} = 546 text{K}$ is correct so it is not the answer for this item.
To convert $text{K}$ to $^circ text{C}$, we use the equation
begin{equation} tag{2}
T_text{C} = T_text{K} – 273
end{equation}
For Option C, we plug in $T_text{K} = 298 text{K}$ to Equation (2)
$$
T_text{C} = 298 – 273
$$
$$
T_text{C} = 25 ^circ text{C}
$$
Thus, $boxed{text{C}. 298 text{K} = 546 ^circ text{C}}$ is incorrect thus it is the answer we are looking for.
For Option D, we plug in $T_text{K} = 88 text{K}$ to Equation (2)
$$
T_text{C} = 88 – 273
$$
$$
T_text{C} = -185 ^circ text{C}
$$
Thus, $text{D}. 88 text{K} = -185 ^circ text{C}$ is correct so it is not the answer for this item.
text{C}. 298 text{K} = 571 ^circ text{C}
$$
Entropy is given by the equation
$$
Delta S = frac{Q}{T}
$$
text{A}. frac{text{J}}{text{K}}
$$
When two objects are in thermal equilibrium, their temperatures are equal.
Heat engines generate energy from the heat transfer between hot and cold reservoirs.
textbf{Given:} \
$m = 87 text{g} = 0.087 text{kg}$ \
$T_i = 14 text{K}$ \
$T_f = 340 text{K}$ \
$T_text{melt} = -97.6 ^circ text{C}$ \
$T_text{boil} = 64.6 ^circ text{C}$ \
$C_text{methanol} = 2450 frac{text{J}}{text{kg} cdot text{K}}$ \
$H_f = 1.09 times 10^5 frac{text{J}}{text{kg}}$ \
$H_v = 8.78 times 10^5 frac{text{J}}{text{kg}}$ \
textbf{Calculation:}\
To have standardized units, we convert the unit of $T_text{melt}$ and $T_text{boil}$ from $^circ text{C}$ to $text{K}$
$$
T_text{K} = T_text{C} + 273
$$
Plugging $T_text{melt}$ into the equation above
$$
T_text{melt} = (-97.6) + 273
$$
$$
T_text{melt} = 175.4 text{K}
$$
Plugging $T_text{boil}$ into the equation above
$$
T_text{boil} = (64.6) + 273
$$
$$
T_text{boil} = 337.6 text{K}
$$
To solve for the heat required to increase the temperature of methanol from $T_i$ to $T_text{melt}$, we use the equation below
$$
Q_1 = mC_text{methanol}(T_text{melt} – T_i)
$$
Plugging in the given values, we have
$$
Q_1 = (0.087)(2450)(175.4 – 14)
$$
$$
Q_1 = 34402.41 text{J}
$$
To solve for the heat required to melt solid methanol, we use the equation below
$$
Q_2 = mH_f
$$
Plugging in the given values, we have
$$
Q_2 = (0.087) cdot (1.09 times 10^5)
$$
$$
Q_2 = 9384 text{J}
$$
To solve for the heat required to increase the temperature of methanol from $T_text{melt}$ to $T_text{boil}$, we use the equation below
$$
Q_3 = mC_text{methanol}(T_text{boil} – T_text{melt})
$$
Plugging in the given values, we have
$$
Q_3 = (0.087) cdot (2450) cdot (337.6 – 175.4)
$$
$$
Q_3 = 34572.93 text{J}
$$
To solve for the heat required to vaporize methanol, we use the equation below
$$
Q_4 = mH_v
$$
Plugging in the given values, we have
$$
Q_4 = (0.087) cdot (8.87 times 10^5)
$$
$$
Q_4 = 77169 text{J}
$$
We solve for the total heat as follows
$$
Q_text{total} = Q_1 + Q_2 + Q_3 + Q_4
$$
$$
Q_text{total} = 34402.41 + 9384 + 34572.93 + 77169
$$
$$
boxed{Q_text{total} = 1.5 times 10^2 text{kJ}}
$$
Q_text{total} = 1.5 times 10^2 text{kJ}
$$
Delta H=363times10^{-3}timesunderbrace{4180}_{text{specific heat}}times14=21.2kJ
$$
21.2kJ
$$
textbf{Given:} \
$m = 81 text{g} = 0.081 text{kg}$ \
$T_i = 0.0 ^circ text{C}$ \
$T_f = 10.0 ^circ text{C}$ \
$H_f = 3.34 times 10^5 frac{text{J}}{text{kg}}$ \
$C = 4180 frac{text{J}}{text{kg} cdot text{K}}$ \
textbf{Calculation:}\
To solve for the heat required to melt ice is given by the equation below
$$
Q_1 = mH_f
$$
Plugging in the given values, we have
$$
Q_1 = (0.081) cdot (3.34 times 10^5)
$$
$$
Q_1 = 27054 text{J}
$$
To solve for the heat required to increase the temperature of water from $T_i$ to $T_f$, we use the equation below
$$
Q_2 = mC(T_f – T_i)
$$
Plugging in the given values, we have
$$
Q_2 = (0.081) cdot (4180) cdot (10 – 0)
$$
$$
Q_2 = 3385.8 text{J}
$$
We can now solve for the total heat as follows
$$
Q_text{total} = Q_1 + Q_2
$$
Plugging in the calculated $Q_1$ and $Q_2$
$$
Q_text{total} = (27054) + (3385.8)
$$
$$
Q_text{total} = 30.4 text{kJ}
$$
Thus, the answer is $boxed{text{C.} 30 text{kJ}}$
text{C.} 30 text{kJ}
$$
textbf{Given:} \
$Q_text{stir} = 0.050 text{J}$ \
$n = 85 text{times}$ \
$T_text{C} = 65 ^circ text{C}$ \
textbf{Calculation:}\
To solve for the total heat generated by the stirring, we use the equation below
$$
Q_text{total} = n cdot Q_text{stir}
$$
Plugging in the given values, we have
$$
Q_text{total} = (85) cdot (0.050)
$$
$$
Q_text{total} = 4.25 text{J}
$$
To convert the unit of temperature from Celsius to Kelvin, we use the equation below
$$
T_text{K} = T_text{C} + 273
$$
Plugging in the given $T_text{C}$, we have
$$
T_text{K} = 65 + 273
$$
$$
T_text{K} = 338 text{K}
$$
To solve for the total increase in entropy, we use the equation below
$$
Delta S = frac{Q_text{total}}{T_text{K}}
$$
Plugging in the calculated $Q_text{total}$ and $T_text{K}$
$$
Delta S = frac{4.25}{338}
$$
$$
Delta S = 0.013 frac{text{J}}{text{K}}
$$
Thus, the answer is $boxed{text{A.} 0.013 frac{text{J}}{text{K}}}$
text{A.} 0.013 frac{text{J}}{text{K}}
$$
textbf{Given:} \
$m = 454 text{g} = 0.454 text{kg}$ \
$T_text{melt} = 0.00 ^circ text{C}$ \
$T_text{boil} = 100.0 ^circ text{C}$ \
$H_f = 3.34 times 10^5 frac{text{J}}{text{kg}}$ \
$H_v = 2.26 times 10^6 frac{text{J}}{text{kg}}$ \
textbf{Calculation:}\
To solve for the heat required to melt ice is given by the equation below
$$
Q_text{melting} = mH_f
$$
Plugging in the given values, we have
$$
Q_text{melting} = (0.454) cdot (3.34 times 10^5)
$$
$$
Q_text{melting} = 1.52 times 10^5 text{J}
$$
To solve for the heat required to vaporize water is given by the equation below
$$
Q_text{steaming} = mH_v
$$
Plugging in the given values, we have
$$
Q_text{steaming} = (0.454) cdot (2.26 times 10^6)
$$
$$
Q_text{steaming} = 1.03 times 10^6 text{J}
$$
To solve for the difference between $Q_text{melting}$ and $Q_text{steaming}$, we use the equation below
$$
Delta Q = Q_text{steaming} – Q_text{melting}
$$
$$
Delta Q = 1.03 times 10^6 – 1.52 times 10^5
$$
$$
boxed{Delta Q = 8.78 times 10^5 text{J}}
$$
To solve for the heat required to increase the temperature of water from $T_text{melt}$ to $T_text{boil}$, we use the equation below
$$
Q_text{heating} = mC(T_text{boil} – T_text{melt})
$$
Plugging in the given values, we have
$$
Q_text{heating} = (0.454) cdot (4180) cdot (100 – 0)
$$
$$
Q_text{heating} = 1.90 times 10^5 text{J}
$$
Thus, we can say that $boxed{Delta Q text{is larger than} Q_text{heating}}$
