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.