Lead couscous weighing 200 g at a temperature of 327 ° C was transferred an amount

Lead couscous weighing 200 g at a temperature of 327 ° C was transferred an amount of heat equal to 5.3 kJ What state is the lead and whether its temperature has increased.

Data: m (mass of a piece of lead) = 200 g = 0.2 kg; t (temperature at which the lead was located) = 327 ºС; Q (the amount of heat transferred to the lead) = 5.3 kJ (5300 J).

Constants: λ (beats heat of fusion) = 0.25 * 105 J / kg; C (specific heat capacity) = 140 J / (kg * K).

1) The amount of heat required to melt lead: Q1 = λ * m = 0.25 * 105 * 0.2 = 5000 J.

2) Change in temperature of liquid lead: Q2 = Q – Q1 = C * m * (t1 – t) and t1 = Q / (C * m) + t = 300 / (140 * 0.2) + 327 = 337.7 ºС.