Detailed Answer

We have the equation $c = \frac{Q}{m\Delta T}$ from which we can see $c$, and $\Delta T$ are constants, meaning if we are going to half the mass, then we will also need to half the energy, to keep the equality.

Detailed Answer

For this, we need to use the equation $L = \frac{Q}{m}$, so we have $Q = 0.25 \cdot 3 \cdot 10^5$, which results in A.

Detailed Answer

We know the power delivered is the energy delivered per unit time, so we have $P = \frac{mc\Delta T}{\Delta t}$. But since we have the graph, we can use its gradient, which is, in fact, $\frac{\Delta T}{\Delta t}$, which we can substitute into our equation, resulting in $P = mc \cdot \text{gradient}$.

Detailed Answer

Since we are only interested in the change in temperature, whether or not we add a constant scalar to both values will not change difference between them.

Detailed Answer

By definition.

Detailed Answer

The heat energy required to change the temperature is given by: $Q = mc\Delta T$

Rearranging to find $\Delta T$: $$\Delta T = \frac{Q}{mc}$$

$$\Delta T = \frac{800}{(2)(4186)} = 0.0956°C$$

The final temperature will be: $$T_{\text{final}} = 10°C + 0.0956°C = 10.1°C$$

Detailed Answer

The apparent brightness is given by the formula: $b = \frac{L}{4\pi d^2}$ where $L = 3.2 \times 10^{26} \, \text{W}$ and $d = 2.0 \times 10^{16} \, \text{m}$.

$$b = \frac{3.2 \times 10^{26}}{4\pi (2.0 \times 10^{16})^2} = 6.4 \times 10^{-8} \, \text{W/m}^2$$

Detailed Answer

The apparent brightness is related to the luminosity and the distance by the formula: $b = \frac{L}{4\pi d^2}$

If the luminosity is doubled ($L' = 2L$) and the distance is halved ($d' = \frac{d}{2}$), the new apparent brightness is:

$$b' = \frac{2L}{4\pi \left( \frac{d}{2} \right)^2} = \frac{2L}{4\pi \cdot \frac{d^2}{4}} = 4 \cdot \frac{L}{4\pi d^2} = 4b$$

Detailed Answer

Wien’s Law is given by: $\lambda_{\text{max}} = \frac{b}{T}$ where $b = 2.9 \times 10^{-3} \, \text{m} \cdot \text{K}$ is the Wien’s displacement constant, and $\lambda_{\text{max}} = 5.8 \cdot 10^{-7} \, \text{m}$. Rearranging the formula to solve for temperature:

$$T = \frac{b}{\lambda_{\text{max}}} = \frac{2.9 \times 10^{-3}}{5.8 \times 10^{-7}} = 5000 \, \text{K}$$

Detailed Answer

The Stefan-Boltzmann law is given by: $P = \sigma A T^4$ where $\sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4$ is the Stefan-Boltzmann constant, $A = 4 \, \text{m}^2$ is the surface area, and $T = 4000 \, \text{K}$. Substituting the values:

$$P = (5.67 \times 10^{-8})(4)(4000)^4 = 5.8 \times 10^7 \, \text{W}$$

Detailed Answer

Using Wien's Law: $\lambda_{\text{max}} = \frac{b}{T}$ where $b = 2.9 \times 10^{-3} \, \text{m} \cdot \text{K}$ and $T = 6000 \, \text{K}$. Substituting the values:

$$\lambda_{\text{max}} = \frac{2.9 \times 10^{-3}}{6000} = 4.8 \times 10^{-7} \, \text{m}$$

Detailed Answer

From the Stefan-Boltzmann law: $$P \propto T^4$$

If the temperature doubles, the new power is: $$P' = P \left( \frac{T'}{T} \right)^4 = P (2)^4 = 16P$$

Detailed Answer

The peak wavelength $\lambda_{\text{max}}$ is inversely proportional to the temperature, given by: $$\lambda_{\text{max}} = \frac{b}{T}$$

Where $b = 2.9 \times 10^{-3}$.

For Star A: $$\lambda_{\text{max, A}} = \frac{2.9 \times 10^{-3}}{6000} = 4.83 \times 10^{-7} \, \text{m}$$

For Star B: $$\lambda_{\text{max, B}} = \frac{2.9 \times 10^{-3}}{8000} = 3.625 \times 10^{-7} \, \text{m}$$

Since Star B has a higher temperature, it emits radiation with a shorter peak wave- length than Star A. Hence, Star B has a shorter peak wavelength.

The total power radiated by a star is given by: $$P = \sigma A T^4$$

Where $A = 4\pi r^2$ is the surface area and $T$ is the temperature.

For Star A: $$A_A = 4 \pi (7 \times 10^8)^2 = 6.16 \times 10^{18} \, \text{m}^2$$ $$P_A = (5.67 \times 10^{-8})(6.16 \times 10^{18})(6000)^4 = 4.52 \times 10^{26} \, \text{W}$$

For Star B: $$A_B = 4 \pi (3 \times 10^8)^2 = 1.13 \times 10^{18} \, \text{m}^2$$ $$P_B = (5.67 \times 10^{-8})(1.13 \times 10^{18})(8000)^4 = 2.62 \times 10^{26} \, \text{W}$$

Therefore, Star A radiates more power than Star B.