A steel kettle weighing 500 g at a temperature of 20 degrees was poured 500 g of hot water

A steel kettle weighing 500 g at a temperature of 20 degrees was poured 500 g of hot water at a temperature of 90 degrees. How will the water temperature in the kettle?

mh = 500 g = 0.5 kg.

mw = 500 g = 0.5 kg.

tch = 20 “С.

tв = 90 “С.

Cst = 500 J / kg * “C.

Cw = 4200 J / kg * “C.

tv “-?

When pouring water, it will give the kettle a certain amount of heat, the kettle will heat up to the temperature tv “, and the kettle will cool down to the temperature tv”.

Let us find the amount of heat Q that the water will give: Qw = Cw * mw * (tv – tw “), where Cw is the specific heat capacity of water, mw is the mass of water.
Let us find the amount of heat Q that the kettle will receive: Qh = Cst * mh * (tv “- tch), where Cst is the specific heat capacity of water, mw is the mass of water.
Let’s write down the heat balance equation: Qw = Qh.
Cw * mw * (tv – tv “) = Cst * mh * (tv” – tch);

Cw * mw * tv – Cw * mw * tv “= Cst * mh * tw” – Cst * mh * tch;

Cw * mw * tv + Cst * mh * tch = Cst * mh * tv “+ Cst * mw * tv”;

Cw * mw * tv + Cst * mh * tch = (Cst * mh + Cw * mw) * tv “;

tv “= (Cw * mw * tv + Cst * mh * th) / (Cst * mh + Cw * mw).

tv “= (4200 J / kg *” C * 0.5 kg * 90 “C + 500 J / kg *” C * 0.5 kg * 20 “C) / (500 J / kg *” C * 0, 5 kg + 4200 J / kg * “C * 0.5 kg) = 82.5” C.

Answer: the water temperature will be set tв “= 82.5” С.

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