How to Calculate Specific Heat and Heat Capacity of Soil?

This article will guide you about how to calculate specific heat and heat capacity of soil.

Specific Heat:

Specific heat may be defined as the amount of heat required to raise the temperature of one gram of a substance by 1°C.

The specific heat of dry soil (0.2 cal/g) is only about one-fifth that of water (1 cal/g). Hence, moist soils are cooler due to their high specific heat and also due to the heat energy spent in evaporation of soil moisture.

One can calculate the specific heat of a soil C_s from the summation of the specific heat times the mass of the individual constituents:

Cs = C₁M₁ + C₂M₂ + C₃M₃ + C₄M₄+ ….C_nM_n [cal/g(°C)

where C_s is specific heat of soil.

M₁, M₂, M₃, M₄ and C₁, C₂….C_nM_n mass and specific heat of individual constituents respectively.

Heat Capacity:

The heat capacity of a given materials is equal to its specific heat multiplied by its mass. The specific heats of soil vary with various soil constituents. It is found that quartz has the lowest specific heat of the major soil constituents and humus has the highest, excepting water.

It is also evident that the heat capacity is influenced by humus and water content of the soil. The heat capacity of a soil constituent is equal to its specific heat times, its density.

The heat capacity of the soil per unit volume can be computed by following the equation as:

C_s = X_SC_S + X_WC_W + X_aC_a [cal/c.c. (°C)]

where C_s is the heat capacity of the soil, X_s, X_w, and X_a are the volume fraction of the solid mass, water and air respectively.

C_s, C_w and C_a are the heat capacities of their respective constituents.

Since the solid mass of the soil consists of mineral and organic matter whose heat capacities per unit volume are approximately 0.45 and 0.60 respectively and the value of air component is too small to be significant.

So the above equation can be simplified as follows:

C_s = 0.46 X_m + 0.60X_o + X_w [X_aC_a is too small and neglected]

Where X_m and X_o are the volume fractions of mineral and organic matter respectively.

Problem:

If a soil contains 30 per cent moisture, calculate the specific heat of the soil. [Specific heat of water and soil solid is 1.0 and 0.2 cal/g]

Solution:

Since soil moisture is expressed as gms of water per 100 gms of soil solids, there are in this case 30 gms of water with each 100 gms of soil.

The number of calories required to raise the temperature of 30 gms of water by 1°C is

30 g × 1 cal/g = 30 cal

Again, the number of calories required to raise the temperature of 100 gms of soil solids by 1°C is

100 g × 0.2 cal/g = 20 cal

Thus, a total of (30 + 20) = 50 calories is required to raise the temperature of 130 gms of the moist soil by 1°C. Since we know, the specific heat is the number of calories required to raise the temperature of 1 gm of wet soil by 1°C and in this case,

Specific heat of the soil = 50/130 = 0.38 cal/g.