A steel teaspoon weighing 25 g, which had previously been on the table in the room, was lowered into

A steel teaspoon weighing 25 g, which had previously been on the table in the room, was lowered into a calorimeter containing 200 g of water at a temperature of 89 ° C. After establishing thermal equilibrium, the water in the calorimeter was cooled to 88 ° C. Neglecting the heat loss and heat capacity of the calorimeter, determine what the temperature of the spoon was before it was lowered into the calorimeter.

mk = 200 g = 0.2 kg.

Cw = 4200 J / kg * ° C.

tv = 89 ° C.

tу = 88 ° C.

ml = 25 g = 0.025 kg.

Cc = 500 J / kg * ° C.

t -?

Since the spoon was lying on the table, it means that it had an air temperature.

Let’s write down the heat balance equation: Qw = Ql.

Qw = Cw * mw * (tv – tу).

Ql = Cs * ml * (tу – t).

Cw * mw * (tv – tу) = Cс * ml * (tу – t).

tу – t = Cw * mw * (tv – tу) / Cs * ml.

t = tу – Cw * mw * (tv – tу) / Cs * ml.

t = 88 ° C – 4200 J / kg * ° C * 0.2 kg * (89 ° C – 88 ° C) / 500 J / kg * ° C * 0.025 kg = 88 ° C – 67.2 ° C = 20.8 ° C.

Answer: the steel spoon was at a temperature of t = 20.8 ° C.