If 11 kJ of heat is transferred to 2300 g of soil at 21°C, what would the new temperature of the soil be?

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

the answer is:  

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