0.2 kg of water was poured into a glass beaker weighing 0.12 kg at a temperature of 15 ° C at a temperature of 100 ° C. At what temperature will thermal equilibrium be established?
The amount of heat possessed by a body at a certain temperature,
is determined by the formula: Q = c * m * t, where Q is the amount of heat (thermal energy) of the body, c is the specific heat capacity of its substance (in our case, c glass = 840 J / kg ° C, c water =
4200 J / kg ° С), m – body weight (in our case, the mass of the glass is 0.12 kg, and the mass of water is 0.2 kg, since the specific density of water is 1000 kg / m³ or 1 kg / dm³ or 1 kg / l), t – the temperature of the water and glass is known (respectively – 100 ° С and 15 ° С).
In the resulting combination (water in a glass), the thermal energies are summed up:
Q = Q glass + Q water.
Substituting the values we know, we get the following:
Q = 840 J / kg ° С * 0.12 kg * 15 ° С + 4200 J / kg ° С * 0.2 kg * 100 ° С. Q = 85512 J.
This energy is possessed by our steam – a glass and water in it.
The average heat capacity of this pair can be calculated based on the values we know –
4200 J / kg ° C * 0.2 kg + 840 J / kg ° C * 0.12 kg) / (0.2 kg + 0.12 kg) = 2940 J / kg ° C.
Now, finally, we can find the steady-state temperature of our pair (the glass is heated, the water has cooled, their temperature is the same):
Q pair = c average * m glass and water * t set.
Solving a simple equation according to the above formula, making the necessary reductions and simplifications, we get –
85512 J = 2940 J / kg ° C * 0.32 kg * t;
t = 85512 J / 2640 J / kg ° С * 0.32 kg = 90.89 ° С (rounding off the infinite fraction to two decimal places).
So, the steady-state temperature of water and a glass, according to the conditions of our example, will be equal to 90.89 ° C.