A snake is sunning itself on a rock.
(a) If the snake is 0.5 m long, what is the approximate amount of radiation it absorbs from the Sun in 11 min? (Assume the snake has a cylindrical shape and ignore both ends, a radius of 0.03 m, and a mass of 2.1 kg. Assume also that the emissivity is 0.5.) Estimate the temperature in the Sun on the rock.
°C
Calculate the amount of radiation the snake absorbs from the Sun in 11 min. (Use your estimates. Note that only half of the snake will face the Sun.)
J
(b) If no heat flows out of the snake via conduction or by evaporative cooling, how much does its temperature increase during 11 min? Assume all the Sun's energy goes into heating the snake and the specific heat of the snake is equal to the specific heat of water. (Use the exact values you enter in previous answer(s) to make later calculation(s).)
K
To answer these questions, we need to use the principles of thermal radiation and heat transfer.
(a) To calculate the amount of radiation the snake absorbs from the Sun in 11 min, we can use the formula for radiant heat transfer:
Q = εσA(T^4 - Ts^4)Δt
Where:
- Q is the amount of radiation absorbed (in Joules),
- ε is the emissivity (a dimensionless quantity that represents the efficiency of an object in emitting and absorbing thermal radiation),
- σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/(m^2 K^4)),
- A is the surface area in contact with the Sun (which is approximately equal to the area of the snake's body),
- T is the temperature of the snake's body (in Kelvin), and
- Ts is the temperature of the Sun (also in Kelvin).
First, we need to calculate the surface area of the snake's body. Since the snake is cylindrical and we are ignoring both ends, the surface area is given by:
A = 2πrh
- π is the mathematical constant pi (approximately equal to 3.14),
- r is the radius of the snake's body (0.03 m),
- h is the height or length of the snake (0.5 m).
Thus, the surface area (A) is:
A = 2π(0.03)(0.5) = 0.03π m^2
Next, let's calculate the temperature of the snake's body (T). We can use the following relationship between mass (m), specific heat (c), and temperature change (ΔT):
Q = mcΔT
Since the specific heat of water is given as an estimate, we can assume the specific heat of the snake is also the same (approximately 4186 J/kg·K). Rearranging the equation, we get:
ΔT = Q / (mc)
Next, let's calculate the temperature of the Sun (Ts). We can use the Stefan-Boltzmann law:
L = 4πR^2σTs^4
Where:
- L is the luminosity of the Sun (approximately 3.8 x 10^26 W),
- R is the radius of the Sun (approximately 6.96 x 10^8 m),
- σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/(m^2 K^4)),
- Ts is the temperature of the Sun (in Kelvin).
Rearranging the equation, we get:
Ts^4 = L / (4πR^2σ)
Now, we can substitute the known values into the formulas and solve for the answers.
(b) To calculate the change in temperature of the snake's body during 11 min, we can use the formula:
ΔT = Q / (mc)
Where:
- ΔT is the change in temperature (in Kelvin),
- Q is the amount of radiation absorbed (in Joules),
- m is the mass of the snake (2.1 kg),
- c is the specific heat (approximately 4186 J/kg·K).
We can take the Q value calculated in part (a) and substitute it into this formula to find the change in temperature.