Solutions 239
where T is the required temperature of the enve-
lope of the station. Finally, we obtain
Tx=V2T 600 K.
2.36. It follows from the graph (see Fig. 67) that
during the first 50 minutes the temperature of the
mixture does not change and is equal to 0 °C.
The amount of heat received by the mixture from
the room during this time is spent to melt ice. In
50 minutes, the whole ice melts and the tempera-
ture of water begins to rise. In 10 minutes (from
50 min to T2 = 60 min), the temperature
increases by AT = 2 °C. The heat supplied to the
water from the room during this time is q =
cwmw AT = 84 kJ. Therefore, the amount of heat
received by the mixture from the room during the
first 50 minutes is Q = 5q = 420 kJ. This amount
of heat was spent for melting ice of mass mice: Q
Xmice• Thus, the mass of the ice contained in the
bucket brought in the room ismice=^ ,•-• 1 2 kff
1C13— - • •2.37*. We denote by a the proportionality factor
between the power dissipated in the resistor and
the temperature difference between the resistor
and the ambient. Since the resistance of the resistor
is R 1 at Ts = 80 °C, and the voltage across it is
U 1 , the dissipated power is UVRi , and we can
writeR
=cc (T 3 — Te).^ (1)The temperature of the resistor rises with the
applied voltage since the heat liberated by the
current increases. As the temperature becomes equal
to T 1 = 100 °C, the resistance of the resistor abrupt-
ly increases twofold. The heat liberated in it will
decrease, and if the voltage is not very high, the
heat removal turns out to be more rapid than the
liberation of heat. This leads to a temperature
drop to 7' 2 = 99 °C, which will cause an abrupt