Answer:
26.98°C
Explanation:
Applying,
Q = cm(t₂-t₁)................ Equation 1
Where Q = Heat transferred to the soil, c = specific heat capacity of dry soil, m = mass of the soil, t₂ = final temperature, t₁ = initial temperature.
From the question,
Given: Q = 11 kJ = 11000 J, m = 2300 g = 2.3 kg, t₁ = 21°C
Constant: c = 800 J/kg.°C
Substitute these values into equation 1
11000 = 800×2.3(t₂-21)
Solving for t₂
t₂-21 = 11000/(800×2.3)
t₂-21 = 11000/1840
t₂-21 = 5.98
t₂ = 5.98+21
t₂ = 26.98°C
