A 2.0 imes 10⁻² ext kg ice at 0^ exto extC is dropped into a vacuum bottle originally holding 0.4 ext kg of water at 35^ exto extC. Assuming negligible heat loss or gain by the bottle, determine (a)

The heat required to melt the ice is given by $Q_{\text{melt}} = m_{\text{ice}} L_f$.

The heat required to melt the ice is $\boxed{6.68 \times 10^3 \text{ J}}$.

Heat gained by ice (to melt and then warm up as water) = Heat lost by initial water.
Let $T_f$ be the final equilibrium temperature.
Heat gained by ice: $Q_{\text{gained}} = Q_{\text{melt}} + m_{\text{ice}} c_{\text{water}} (T_f - 0^\circ\text{C})$
Heat lost by water: $Q_{\text{lost}} = m_{\text{water}} c_{\text{water}} (35^\circ\text{C} - T_f)$

Combine terms with $T_f$:

The final temperature after thermal equilibrium is attained is approximately $\boxed{29.53^\circ\text{C}}$.

7. What is the absolute temperature of boiling water and melting ice?

The absolute temperature (in Kelvin) is found by adding 273.15 to the Celsius temperature: $T_K = T_C + 273.15$.

Melting ice is at $0^\circ\text{C}$.

Boiling water is at $100^\circ\text{C}$.

The absolute temperature of melting ice is $\boxed{273.15 \text{ K}}$ and boiling water is $\boxed{373.15 \text{ K}}$.

8. Determine the temperature whose Fahrenheit and Kelvin scales have the same reading.

Let the temperature be $x$. So, $T_F = x$ and $T_K = x$.
The conversion from Celsius to Fahrenheit is $T_F = \frac{9}{5}T_C + 32$.
The conversion from Celsius to Kelvin is $T_K = T_C + 273.15$, which means $T_C = T_K - 273.15$.

Subtract $1.8x$ from both sides:

The temperature at which Fahrenheit and Kelvin scales have the same reading is approximately $\boxed{574.59}$.

9. A scientist uses a $\gamma$ scale for measuring temperature. In this scale water melts at $10^\circ\gamma$, and boils at $130^\circ\gamma$. The scientist measures the temperature at which sodium melts to be $127^\circ\gamma$. Express this temperature in Kelvin.

Water melts at $0^\circ\text{C}$ or $273.15 \text{ K}$.
Water boils at $100^\circ\text{C}$ or $373.15 \text{ K}$.
For the $\gamma$ scale: $T_{\gamma, \text{melt}} = 10^\circ\gamma$, $T_{\gamma, \text{boil}} = 130^\circ\gamma$.
For the Kelvin scale: $T_{K, \text{melt}} = 273.15 \text{ K}$, $T_{K, \text{boil}} = 373.15 \text{ K}$.

The general formula for converting between two linear temperature scales is:

So, we can write:

The temperature of sodium melting is $T_\gamma = 127^\circ\gamma$.

Multiply both sides by 100:

Add 273.15 to both sides:

The melting temperature of sodium in Kelvin is $\boxed{370.65 \text{ K}}$.