Съдържание

IZhO 2006, Теория — Задача 2

Оригинал на условието (Word)Съобщи за грешка

Всички материали от IZhO 2006

2nd INTERNATIONAL ZHAUTYKOV PHYSICS AND MATHEMATICS OLYMPIAD. KAZAKHSTAN, ALMATY 2006. Physics, Theoretical competition. · 8 т.

Условие

The stove is used to heat the room, and the temperature in the room is $T_1 = 17^{\circ}\mathrm{C}$ , while the temperature outside is $T_{\mathrm{o}} = 7^{\circ}\mathrm{C}$ . It is suggested to use the ideal thermal pump instead, using the reversed Carnot cycle. Degree of efficiency of the engine, used in reverse Carnot cycle is $\eta = 60\%$ . Assuming that the heat exchange between the room and the outside is proportional to the temperature difference, and that the engine uses the same amount of fuel per unit time as the stove, calculate the temperature in the room, if

1) engine is located outside of the room

2) engine is located inside the room.

Отговори

Решение

Let the heat transmission factor be k, then the energy losses of the stove are given by

A = k(T_1 - T_0),

(1.0 point)

where A – is the energy consumed by the stove in a unit time, $T = t^{0}\mathrm{C} + 273$ . The thermal pump operates using the reversed Carnot cycle, with maximal temperature being equal to the room temperature, and the minimal – to the temperature outside.

Denote the room temperature by T, then the efficiency of the Carnot cycle would be $(T - T_0)/T$ , and in the reversed Carnot cycle the following equality will hold

\frac{A_{Carnot}}{Q_2} = \frac{T - T_0}{T},

(2 points)

c) where $Q_2$ is the heat transferred to the room in a unit time,

A_{Carnot} = \eta A,

(1.0 point)

where A is the work performed by the engine in a unit time. The energy balance in the case of engine located outside of the room is

k(T - T_0) = Q_2 = \frac{TA_{Carnot}}{T - T_0} = \frac{\eta T}{T - T_0}k(T_1 - T_0) \Rightarrow T - T_0 = \frac{T}{T - T_0}\eta(T_1 - T_0),

(1,5 points)

whence we find

(T - T_0)^2 - \eta(T_1 - T_0)(T - T_0) - \eta T_0(T_1 - T_0) = 0.

Solving this equation we find $T - T_0 = 44.1^{0}$ , hence the temperature in the room would be

$t_2 = 54{,}1^{0}\mathrm{C}$ . (1.0 point)

When the engine is inside the room an additional energy equal to $(1 - \eta)A$ is dissipated in the room, thus the energy balance equation

k(T - T_0) = \frac{T\eta}{T - T_0}k(T_1 - T_0) + (1 - \eta)(T_1 - T_0)k,

(1.0 point)

hence

(T - T_0)^2 - \eta(T_1 - T_0)(T - T_0) - T_0(T_1 - T_0) = 0.

The positive solution of that equation is $T - T_0 = 56^{0}$ and thus the temperature established in the room will be

$t_3 = 66^{0}\mathrm{C}$ (0,5 points)

Оригинал в Архива: IZhO-2006_╨óheory_eng.doc · официални решения: IZhO-2006_╨óheory_eng_sol.doc