Santosh,

Yes.

You know the incident intensity of the sunlight, the duratino of exposure, and you know the area that the sunlight strikes (the projected area of the car's interior). Then you need to know the albedo and the infra-red emissivity.

(polished surfaces have a low emissivity, and car interiors tend to be dark = low albedo)

You have everything you need.

Hello Santosh

i think this is helpful to u

Write down the following formula: the amount of heat transferred or absorbed Q = mass (m) x change in temperature (T) x specific heat (c). The SI units for heat transferred is joules; however, calories are still often used for problems involving water. The conversion from joules to calories is 4.18 J = 1 cal.

Determine what the specific heat c is for the object in question. Specific heat capacity or specific heat is the amount of heat it takes to raise the temperature of a substance by one degree Celsius. This quantity can be looked up from a list of specific heat values. The units for specific heat are joules per gram times degrees Celsius (J/g * °C) or calories per gram times degrees Celsius (cal/g * ° C).

Determine the change in temperature of the object caused by the absorbed heat. This can be done by taking the initial temperature T(i) of the object and subtracting it from the final temperature T(f) of the object after the heat is absorbed. For SI units temperature is usually given in degrees Celsius. For example, if an object is 10° C and it changes to 40° C after the heat absorption, then the change in temperature T = T(f) - T(i) = 40° C - 10° C = 30° C.

Weigh the object to determine its mass using a balance. It is important to record an accurate reading. The mass will affect the outcome of the heat absorption. A larger mass will cause a larger outcome for the heat transfer, and a smaller mass will do the opposite. The SI units for mass will be in grams.

Substitute all of the known values into the original equation and solve it using a calculator to determine the amount of heat absorbed by a substance. For example, if 250 grams of water have a change in temperature of 30° C and the specific heat value is determined to be 4.184 J/g * °C, then the heat absorption for the water will be Q = m * c * T = 250 g * 4.184 (Joules per calorie) * 30° C = 31,380 Joules.

Anurag,

I am puzzled by your reference to water - Santosh needs to (surely) know more about the typical albedo of car interior fabrics and the emissivity of polished car paint than he needs to know about the heat capacity of water!

Santosh.

The power delivered will scale as projected area (A) and the intensity of sunlight (I).

You must also consider the heat loss mechanisms, which will be from convection from the exterior of the car's passenger shell.

Garry,

Thank you for your answer and time. Very helpful.

Could you elaborate a little more on 'albedo' in this context here? 

Will look forward to hear. Thanks in advance.

Hi Anurag,

Thanks for your time. Gave me a perspective. Your answer was helpful too.

Santosh,

The albedo of a material's surface finish is a measure of how much light is scattered back from that material, as a ratio of the incoming light.

So, a perfectly white matt surface has an albedo of 1 (all the incoming light is scattered perfectly - one assumes a hemispherical 'collector' to sample the scattered light)

Car materials tend to be both diffusely scattering (soft, and textured) and dark (hide the stains and spills!)

The energy absorbed by a material will vary with (1-a), for an albedo a.

So: incident sunlight, at 500 W m^-2, hitting an area of 1m^2, with an albedo of 0.2, absorbs 500 x 1 x (1-0.2) joules per second.

A far more complicated understanding is needed for the emissivity! Bear in mind, at some point you have an equilibrium between the incoming heat rate and the cooling rate (car interiors do not warm without limit - they reach a peak and stay there - and finding that value is not trivial).

Garry,

I couldn't thank you more for your time and information, appreciate it.

Is there any study material or literature work could you share with me? I want to dig deeper in this. I need to find out exactly how warm it gets in the inside of a car. Any leads will be of great help. 

Thank you once again.

Santosh,

The problem divides into two halves.

There's the warming phase (in which we neglect losses), and then there's an equilibrium phase (where heat loss balances heat gain).

In the first part, you know the total incident warming power that falls on the car.

Q (W) = projected area (m^2). Intensity (W/m^2)

A fraction of that projected area, ref_frac, is shiny painted metal with a reflectivity R.

Q_ref = ref_frac . projected_area (m^2) . R . Intensity (W/m^2)

This is the reflected intensity.

The power absorbed by the shiny paintwork, Q_abs_metal, is;

Q_abs_metal = projected_area (m^2) . (1-R) . Intensity (W/m^2)

There's also another component to the absorbed light: the light that passes through the windows, and is absorbed by the interior materials.

If ref_frac is the fractional area occupied by paintwork, then (1-ref_frac) is the fractional area that is window (to first order).

So with an albedo, a (the Bond albedo, IIRC), the heating power absorbed by the soft-furnishings is:

Q_soft = (1-ref_frac). projected area (m^2) . Intensity (W/m^2) . (1-a)

Add the two together, Q_soft+Q_abs_metal and that's the total power that warms the car.

Now, do have a think about how power 'in' has to balance power 'out' in equilibrium. The car can lose heat through radiation (badly) or convection.

You've got the power 'in', now consider how convection cools a hot object. That's more complicated, but if you treat the car as a sphere at constant temperature, that's a start.

You also need to consider the "greenhouse effect". Most of the energy in sunlight is in visible wavelengths that will pass through the windscreen and be absorbed by the interior materials. Those surfaces then radiate but in the infrared band. The windscreen glass may have quite different optical properties at the longer wavelength so the heat can be trapped inside.

The signalled reference may be found helpful: Jan F. Kreider, "Medium and High Temperature Solar Processes", Academic Press, New York / London, 1979; Ch. 2: "Principles of solar radiation and optics"; http://www.sciencedirect.com/science/article/pii/B9780124259805500078