A cast iron iron weighing 2 kg and a copper pot weighing 0.5 kg are placed on a stove with a temperature

A cast iron iron weighing 2 kg and a copper pot weighing 0.5 kg are placed on a stove with a temperature of 200 degrees. Which of these bodies heats up faster? Why? Which of them needs less heat for this? How many times?

t = 200 “C.

mу = 2 kg.

mk = 0.5 kg.

Cch = 540 J / kg * “C.

Cm = 400 J / kg * “C.

Qу / Qк -?

To heat the iron from the temperature t0 to the temperature t, the amount of heat Qу is required, which is determined by the formula: Qу = Cч * mу * (t – t0), where Cч is the specific heat capacity of cast iron, mу is the weight of the iron.

Qу = 540 J / kg * “C * 2 kg * (t – t0) = 1080 J /” C * (t – t0).

To heat the pot from temperature t0 to temperature t, the amount of heat Qk is required, which is determined by the formula: Qk = Cm * mk * (t – t0), where Cm is the specific heat capacity of copper, mk is the mass of the pot.

Qk = 400 J / kg * “C * 0.5 kg * (t – t0) = 200 J /” C * (t – t0).

Since (t – t0) is the same for the iron and the bowler hat, then Qу> Qк.

The pot will heat up faster than the iron because the pot needs less heat to heat up than the iron.

Qу / Qк = 1080 J / “C * (t – t0) / 200 J /” C * (t – t0) = 5.4.

Answer: The pot will heat up faster than the iron. Qу / Qк = 5.4.