How much should the temperature of a piece of lead weighing 100 g be increased in order

How much should the temperature of a piece of lead weighing 100 g be increased in order for its internal energy to increase by 280 J?

Data: m (mass of a piece of lead) = 100 g (0.1 kg); ΔQ (change (increase) in the internal energy of a piece of lead) = 280 J.

Constants: С (specific heat capacity of lead) = 140 J / (kg * ºС).

The required temperature rise of a piece of lead is calculated from the formula: ΔQ = C * m * Δt, whence Δt = ΔQ / (C * m).

Let’s calculate: Δt = 280 / (140 * 0.1) = 20 ºС.

Answer: The temperature of a piece of lead needs to be increased by 20 ºС.