A lead body weighing 2 kg, when cooled, emits an amount of heat equal to 2600 J.

A lead body weighing 2 kg, when cooled, emits an amount of heat equal to 2600 J. How many degrees has its temperature decreased?

Problem data: m (mass of the considered lead body) = 2 kg; Q (heat released by the body) = 2600 J.

Reference values: Cc (specific heat of lead) = 140 J / (kg * K).

The decrease in the temperature of the lead body under consideration is calculated by the formula: Q = Cc * m * Δt, whence Δt = Q / (Cc * m).

Let’s calculate: Δt = 2600 / (140 * 2) = 9.29 K.

Answer: The temperature of the lead body in question will decrease by 9.29 K.