Heat Balance The ideal calorimeter contained water at a temperature of 10 ° C. A body with a temperature

Heat Balance The ideal calorimeter contained water at a temperature of 10 ° C. A body with a temperature of 100 ° C was placed in the water. After a while, the calorimeter reached a temperature of 40 ° C. What will be the equilibrium temperature in the calorimeter if one more body of the same with a temperature of 100 ° C is placed in it, without removing the first one?

Task data: tv (initial water temperature) = 100 ºС; tт (temperature of two bodies taken) = 100 ºС; tр (first equilibrium temperature) = 40 ºС.

The first heat balance equation: St * mt * (tt – tp) = Sv * mw * (tr – tb), whence St * mt / Sv * mw = (tr – tb) / (tt – tr).

The second heat balance equation: St * 2 * mt * (tt – tx) = Sv * mw * (tx – tv), whence St * mt / Sv * mw = (tx – tv) / (2 * (tt – tx) ).

The resulting equality: (tp – tb) / (tt – tp) = (tx – tb) / (2 * (tt – tx)).

2 * (tт – tх) * (tр – tв) = (tт – tв) * (tх – tв).

Substitute the known values: 2 * (100 – tx) * (40 – 10) = (100 – 40) * (tx – 10).

2 * (100 – tx) * 30 = 60 * (tx – 10).

60 * (100 – tx) = 60tx – 600.

6000 – 60tx * 30 = 60tx – 600.

120tx = 6600.

tх = 6600/120 = 55 ºС.

Answer: The second equilibrium temperature is 55 ºС.