The rate of energy loss via radiation from a body can be calculated using the StefanBoltzmann law, which is applied to thermal radiation emitted by an object. Here's the formula:
P=σ⋅A⋅ε⋅T4
where:
 ( P ) is the power radiated (energy per unit time, in watts).
 ( \sigma ) is the StefanBoltzmann constant, approximately ( 5.67 \times 10^{8} , \text{W} , \text{m}^{2} , \text{K}^{4} ).
 ( A ) is the surface area of the radiating body in square meters.
 ( \varepsilon ) is the emissivity of the surface (a value between 0 and 1, where 1 represents a perfect blackbody).
 ( T ) is the absolute temperature of the body in Kelvin.
Steps:
 Determine Surface Area (A): Calculate or measure the surface area of the object.
 Determine Temperature (T): Measure the temperature of the object in Kelvin.
 Determine Emissivity (ε): Use the material’s emissivity; typical values can be found in tables based on material type.
 Calculate Power Radiated (P): Substitute values into the StefanBoltzmann formula.
This gives the rate at which energy is radiated. If you want the total energy radiated over a certain time ( t ), multiply ( P ) by ( t ):
[ E = P \cdot t ]
For complex cases (e.g., nonuniform temperature), you might integrate over the surface with varying ( T ) values.
So at what rate would Apollo 13 be losing energy, roughly?
To estimate the energy loss rate via radiation for Apollo 13, we’d need to approximate some of its parameters: its surface area, the temperature it was maintaining, and the emissivity of its materials.
Here’s a rough calculation with these estimates:
Step 1: Estimating the Surface Area (A)
Apollo 13’s Command and Lunar Modules together had an approximate surface area. The Command Module alone, shaped somewhat like a truncated cone, has a surface area of about ( 30 , \text{m}^2 ). Including the Lunar Module, let’s estimate a combined area of approximately ( 50 , \text{m}^2 ).
Step 2: Estimating Temperature (T)
During Apollo 13’s crisis, the interior was reportedly very cold, close to freezing, around ( 4^\circ \text{C} ) (or ( 277 , \text{K} )). However, for the purpose of estimating energy lost to space, we should consider the exterior temperature at the hull. We can estimate this to be about ( 250 , \text{K} ) due to the cold environment of space and minimal internal heating.
Step 3: Emissivity (ε)
The emissivity of the spacecraft's outer shell, which was primarily metal with reflective insulation, would be around ( \varepsilon = 0.5 ) for aluminum and similar materials in space conditions.
Step 4: Calculation
Using the StefanBoltzmann law:
[ P = \sigma \cdot A \cdot \varepsilon \cdot T^4 ]
Substitute the values:
 ( \sigma = 5.67 \times 10^{8} , \text{W} , \text{m}^{2} , \text{K}^{4} )
 ( A = 50 , \text{m}^2 )
 ( \varepsilon = 0.5 )
 ( T = 250 , \text{K} )
Calculating:
[ P = 5.67 \times 10^{8} \times 50 \times 0.5 \times (250)^4 ]
Let's compute this.
The approximate rate of energy loss via radiation from Apollo 13 would be around ( 5537 , \text{W} ) (watts).
This rate would represent a continuous cooling effect, which, combined with the lack of internal power and heating, led to the intense cold experienced by the crew.
OK but we need to remove the side facing the Sun and calculate energy being absorbed don't we?
Yes, you're absolutely correct! When calculating the energy lost to space, we should account for the fact that only part of Apollo 13’s surface was radiating heat away. Additionally, we need to estimate the energy it was absorbing from the Sun on the sunlit side.
Let’s break down the full calculation:
Step 1: Effective Radiating Area
Assuming Apollo 13 had one side roughly facing the Sun, we can approximate that half of its surface area was radiating heat into space. So, let’s use an effective radiating area of:
[ A_{\text{effective}} = \frac{A}{2} = \frac{50}{2} = 25 , \text{m}^2 ]
Step 2: Energy Absorbed from the Sun
To estimate the absorbed energy, we consider:
[ P_{\text{absorbed}} = A_{\text{sunlit}} \cdot \alpha \cdot S ]
where:
 ( A_{\text{sunlit}} ) is the area facing the Sun. Assuming direct sunlight on half the surface, we use ( 25 , \text{m}^2 ).
 ( \alpha ) is the absorptivity, around ( 0.3 ) for a reflective, metallic spacecraft.
 ( S ) is the solar constant (about ( 1361 , \text{W/m}^2 ) at Apollo 13's distance from the Sun).
Now, we’ll calculate both the radiated and absorbed power to find the net energy loss.
For Apollo 13:
 Power radiated from the shaded side: approximately ( 2769 , \text{W} ).
 Power absorbed from the Sun on the sunlit side: approximately ( 10,208 , \text{W} ).
This results in a net energy gain of about ( 7439 , \text{W} ), meaning that Apollo 13 would have absorbed more energy from the Sun than it radiated away. However, in practice, Apollo 13 still experienced intense cooling because the interior heating systems were off, and heat wasn’t effectively transferred from the sunlit areas to the shaded regions where the crew resided.
But they had to keep spinning the craft 30 degrees at intervals to sop it being destroyed
You're right; Apollo 13 was indeed rotated at intervals to prevent one side from overheating while the other side froze. This maneuver, known as a "Passive Thermal Control" (PTC) or "barbecue roll," was crucial for evenly distributing solar heating around the spacecraft, avoiding excessive temperature gradients that could damage the craft.
This rolling motion would have periodically changed the areas exposed to the Sun, slightly adjusting the effective radiating and absorbing surfaces over time, which helped to manage the temperature distribution without active heating systems. It’s a great example of how crucial thermal management is in space, especially in emergencies.
So How Could They Remain in Shade?
For Apollo 13 to remain in the shade as much as possible while balancing temperature, they would rely on the "barbecue roll" to rotate the craft at intervals, presenting each part to sunlight briefly. This maneuver minimizes prolonged exposure of any single area to direct sunlight while ensuring all sides receive some heating, keeping extreme temperatures in check.
The challenge of maintaining shade involves:

Slow and Controlled Rotation: By rotating every 30 degrees at intervals, the spacecraft's orientation toward the Sun changes frequently. This balances out the heating, reducing hotspots while still preventing prolonged exposure to space's intense cold.

Optimal Sun Angle Management: By controlling the angle of rotation, Apollo 13's crew ensured the craft stayed in a position where direct sunlight was only glancing off portions of the surface. This setup let them balance temperature while preventing excessive heating that could occur if one side faced the Sun continuously.

Minimal Surface Area in Direct Sunlight: Since they maintained intervals in the craft's rotation, no side stayed in full sunlight for too long. This method allowed indirect, periodic warming without full exposure, effectively keeping much of the craft in shade relative to intense solar heating.
This delicate balance between sun exposure and shade was crucial to prevent freezing in shaded areas and overheating in sunlit areas, all without active temperature control systems.
Let's recalculate just for the LEM
Let's focus on the Lunar Module (LEM) of Apollo 13 alone. We’ll estimate its energy loss through radiation and its solar energy absorption, taking into account the rotating maneuvers.
Revised Parameters for the LEM

Surface Area (A): The LEM’s surface area was smaller than the combined modules. We can approximate it at around ( 20 , \text{m}^2 ).

Temperature (T): The outer surface temperature during the Apollo 13 crisis could still be around ( 250 , \text{K} ).

Emissivity (ε): Similar to before, we’ll assume ( \varepsilon = 0.5 ) for its metallic surface.

Absorptive Side (A_sunlit): Since we’ll consider only half of the LEM facing the Sun, we’ll use ( 10 , \text{m}^2 ) for the sunlit side.

Absorptivity (α): We’ll keep ( \alpha = 0.3 ).
Let’s recalculate the radiated and absorbed power for the LEM under these conditions.
For the Lunar Module (LEM) of Apollo 13:
 Power radiated from the shaded side: approximately ( 1107 , \text{W} ).
 Power absorbed from the Sun on the sunlit side: approximately ( 4083 , \text{W} ).
This results in a net energy gain of about ( 2976 , \text{W} ). Despite this net gain, the rotating maneuver (barbecue roll) was still essential for managing the internal temperatures, as it distributed this absorbed energy across the LEM more evenly, preventing one side from overheating and the other from freezing.
One side was a 3Kw heater and they were cold?
Yes, the paradox of a "3 kW heater" on one side, yet experiencing intense cold, underscores the unique thermal challenges in space. Despite the LEM absorbing significant energy from sunlight, this energy didn't translate effectively into warmth for the crew inside due to several factors:

Vacuum of Space: Space lacks an atmosphere to conduct or convect heat, so the only way for heat to reach the shaded and interior areas was through conduction within the spacecraft materials. Unfortunately, the LEM’s outer insulation was designed to prevent heat transfer, both in and out, which reduced the transfer of warmth to the interior.

Thermal Isolation: The materials and design of the spacecraft prioritized insulating the interior from both extreme heat and cold. This was effective under normal mission conditions, where internal systems provided heating. However, with Apollo 13’s power systems mostly shut down, the insulation worked against the crew by trapping the cold inside.

Uneven Heating with Sunlight Intervals: The "barbecue roll" allowed the craft to avoid overheating by distributing sunlight exposure across surfaces. While it prevented damage, it also limited how much solar energy could contribute to internal warmth, as no surface was in the Sun long enough to conductively transfer significant heat inward.

Insufficient Internal Power: Normally, electrical systems and equipment in the Command Module generated heat, balancing out any external cold. With those systems offline, there was minimal internal heat generation, making the crew reliant on residual warmth and passive solar exposure.
In essence, while the sunlit side absorbed solar energy, the spacecraft's design for thermal isolation and the lack of active heating systems resulted in much of that energy remaining outside, leaving the crew to endure the freezing cold.