Data:
$a_{\text{g Earth}} = 9.8\,\text{m/s}^2$
Water:
$k_{l\,\text{fusion}} = 334\,000\,\text{J/kg}$
$k_{s\,\text{liquid}} = 4\,184\,\text{J/kg°C}$
$k_{s\,\text{solid}} = 2\,100\,\text{J/kg°C}$
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1. Why even thought carpet and tile floor have been in contact all night we say that the carpet is warmer than the tile floor? (1 point)
Feedback: The carpet has a higher specific heat capacity, meaning it can store more heat energy per unit mass.
2. Indicate the heat transfer methode applied in teh next cases (1 point)
Feedback:
3. A water wave travels at 6 m/s and has a wavelength of 3 m. What is its period (s)? (2 points)
Feedback: **Given:** $v = 6\,\text{m/s}$, $\lambda = 3\,\text{m}$. First find the frequency: $f = \frac{v}{\lambda} = \frac{6\,\text{m/s}}{3\,\text{m}} = 2\,\text{Hz}$. Then find the period: $T = \frac{1}{f} = \frac{1}{2\,\text{Hz}} = 0.5\,\text{s}$. The period is **0.5 s**.
4. A vehicle with an internal combustion engine produces 120 000 J. (4 points)
Feedback:
5. A ball is thrown vertically upward from a height of 2 m above the ground with an initial speed of 20 m/s. Assume that air resistance is negligible. Calculate: (4 points)
Feedback:
6. A student has 200 g of ice at –20°C. If ice is heated until it becomes water at 70°C. Calculate the total energy added (J). (1 point)
Feedback: First we need to use the right units $200\,\text{g} = 0.2\,\text{kg}$. The total energy added is the sum of the energy required to heat the ice from -20°C to 0°C, the energy required to melt the ice at 0°C, and the energy required to heat the water from 0°C to 70°C. So we have: $Q = m c_{s\,\text{solid}} \Delta T + m k_{l\,\text{fusion}} + m c_{s\,\text{liquid}} \Delta T = (0.2)(2100)(20) + (0.2)(334000) + (0.2)(4184)(70) = 8400 + 66800 + 58576 = 108800$ J.
7. A container has 500 g of water at 15°C. If 42 000 J of heat energy is added. Calculate the final temperature of the water (°C). (2 points)
Feedback: First we need to use the right units $500\,\text{g} = 0.5\,\text{kg}$. The energy added is used to heat the water, so we can use the formula $Q = m c_{s\,\text{liquid}} \Delta T$. Rearranging the formula to find the final temperature, we get $\Delta T = \frac{Q}{m c_{s\,\text{liquid}}} = \frac{42000}{0.5 \times 4184} = \frac{42000}{2092} \approx 20.1$°C. Adding this to the initial temperature of 15°C, the final temperature is approximately 35.1°C.
8. A student mixes: 2 kg of water at 80°C and 3 g of water at 20°C. Calculate the final equilibrium temperature. (2 points)
Feedback: First we need to use the right units $3\,\text{g} = 0.003\,\text{kg}$. The heat lost by the hot water will be equal to the heat gained by the cold water. So we can set up the equation: $m_1 c \Delta T_1 = m_2 c \Delta T_2$, where $m_1 = 2$ kg, $m_2 = 0.003$ kg, $c = 4184$ J/kg°C, $\Delta T_1 = 80 - T_f$, and $\Delta T_2 = T_f - 20$. Solving for $T_f$, we get $2 \times 4184 \times (80 - T_f) = 0.003 \times 4184 \times (T_f - 20)$. This simplifies to $668800 - 8368T_f = 12.552T_f - 251.04$. Combining like terms gives us $668800 + 251.04 = 8368T_f + 12.552T_f$, which simplifies to $669051.04 = 8380.552T_f$. Finally, solving for $T_f$ gives us $T_f \approx \frac{669051.04}{8380.552} \approx 79.97$°C.