A 1.0-kg block of aluminum is at a temperature of 50°C. How much thermal energy will it lose when its temperature is reduced by half? The specific heat of aluminum is 897 J/(kg •K).

22425 J

Explanation:

From the question,

Applying

Q = cm(t₂-t₁) Equation 1

Where Q = Thermal Energy, c = specific heat capacity of aluminium, m = mass of aluminium, t₂ = Final Temperature, t₁ = Initial Temperature.

Given: c = 897 J/kg.K, m = 1.0 kg, t₁ = 50 °C, t₂ = 25 °C (The final temperature is reduced by half)

Substitute these values into equation 1

Q = 897×1×(25-50)

Q = 897×(-25)

Q = -22425 J

Hence the thermal energy lost by the aluminium is 22425 J

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