The specific heat of soil is 0.20 kcal/kg*C and the specific heat of water is 1.00 kcal/kg*C. This means that if 1 kg of soil and 1 kg of water each receive 1 kcal of energy, ideally.

Respuesta :

Answer: The soil will be [tex]4\°C[/tex] warmer than the water.

Explanation:

The heat (thermal energy) absorbed can be found using the following equation:

[tex]Q=m.C.\Delta T[/tex]

Where:

[tex]Q[/tex] is the heat  

[tex]m[/tex] is the mass of the element

[tex]C[/tex] is the specific heat capacity of the material.

[tex]\Delta T[/tex] is the variation in temperature

In the case of soil we have:

[tex]Q_{soil}=m_{soil}.C_{soil}.\Delta T_{soil}[/tex] (1)

Where:

[tex]Q_{soil}=1 kcal[/tex]

[tex]m_{soil}=1 kg[/tex]

[tex]C_{soil}=0.2 kcal/kg \°C[/tex]

[tex]\Delta T_{soil}[/tex]

In the case of water we have:

[tex]Q_{water}=m_{water}.C_{water}.\Delta T_{water}[/tex] (2)

Where:

[tex]Q_{water}=1 kcal[/tex]

[tex]m_{water}=1 kg[/tex]

[tex]C_{water}=1 kcal/kg \°C[/tex]

[tex]\Delta T_{water}[/tex]

Isolating [tex]\Delta T[/tex] from both equations:

[tex]\Delta T_{soil}=\frac{Q_{soil}}{m_{soil}.C_{soil}}[/tex] (3)

[tex]\Delta T_{soil}=\frac{1 kcal}{1 kg(0.2 kcal/kg \°C)}[/tex]

[tex]\Delta T_{soil}=5\°C[/tex] (4)

[tex]\Delta T_{water}=\frac{Q_{water}}{m_{water}.C_{water}}[/tex] (5)

[tex]\Delta T_{water}=\frac{1 kcal}{1 kg(1 kcal/kg \°C)}[/tex]

[tex]\Delta T_{water}=1\°C[/tex] (6)

Comparing (4) and (6) we can find the soil will be [tex]4\°C[/tex] warmer than the water.

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