Съдържание

IPhO 1969, Теория — Задача 2

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IPhO 1969

3rd International Physics Olympiad 1969, Brno, Czechoslovakia

Условие

Water of mass $m_2$ is contained in a copper calorimeter of mass $m_1$ . Their common temperature is $t_2$ . A piece of ice of mass $m_3$ and temperature $t_3 < 0\,{}^{\circ}\mathrm{C}$ is dropped into the calorimeter.

a) Determine the temperature and masses of water and ice in the equilibrium state for general values of $m_1$ , $m_2$ , $m_3$ , $t_2$ and $t_3$ . Write equilibrium equations for all possible processes which have to be considered.

b) Find the final temperature and final masses of water and ice for $m_1 = 1.00\ \mathrm{kg}$ , $m_2 = 1.00\ \mathrm{kg}$ , $m_3 = 2.00\ \mathrm{kg}$ , $t_2 = 10\,{}^{\circ}\mathrm{C}$ , $t_3 = -20\,{}^{\circ}\mathrm{C}$ .

Neglect the energy losses, assume the normal barometric pressure. Specific heat of copper is $c_1 = 0.1\ \mathrm{kcal/kg{\cdot}{}^{\circ}C}$ , specific heat of water $c_2 = 1\ \mathrm{kcal/kg{\cdot}{}^{\circ}C}$ , specific heat of ice $c_3 = 0.492\ \mathrm{kcal/kg{\cdot}{}^{\circ}C}$ , latent heat of fusion of ice $l = 78{,}7\ \mathrm{kcal/kg}$ . Take $1\ \mathrm{cal} = 4.2\ \mathrm{J}$ .

Отговори

Решение

We use the following notation:

$t$	temperature of the final equilibrium state,
$t_0 = 0\,{}^{\circ}\mathrm{C}$	the melting point of ice under normal pressure conditions,
$M_2$	final mass of water,
$M_3$	final mass of ice,
$m_2' \le m_2$	mass of water, which freezes to ice,
$m_3' \le m_3$	mass of ice, which melts to water.

a) Generally, four possible processes and corresponding equilibrium states can occur:

$t_0 < t < t_2$ , $m_2' = 0$ , $m_3' = m_3$ , $M_2 = m_2 + m_3$ , $M_3 = 0$ .

Unknown final temperature $t$ can be determined from the equation

(m_1c_1 + m_2c_2)(t_2 - t) = m_3c_3(t_0 - t_3) + m_3l + m_3c_2(t - t_0)\,. \qquad (7)

However, only the solution satisfying the condition $t_0 < t < t_2$ does make physical sense.

$t_3 < t < t_0$ , $m_2' = m_2$ , $m_3' = 0$ , $M_2 = 0$ , $M_3 = m_2 + m_3$ .

Unknown final temperature $t$ can be determined from the equation

m_1c_1(t_2 - t) + m_2c_2(t_2 - t_0) + m_2l + m_2c_3(t_0 - t) = m_3c_3(t - t_3)\,. \qquad (8)

However, only the solution satisfying the condition $t_3 < t < t_0$ does make physical sense.

$t = t_0$ , $m_2' = 0$ , $0 \le m_3' \le m_3$ , $M_2 = m_2 + m_3'$ , $M_3 = m_3 - m_3'$ .

Unknown mass $m_3'$ can be calculated from the equation

(m_1c_1 + m_2c_2)(t_2 - t_0) = m_3c_3(t - t_3) + m_3'l\,. \qquad (9)

However, only the solution satisfying the condition $0 \le m_3' \le m_3$ does make physical sense.

$t = t_0$ , $0 \le m_2' \le m_2$ , $m_3' = 0$ , $M_2 = m_2 - m_2'$ , $M_3 = m_3 + m_2'$ .

Unknown mass $m_2'$ can be calculated from the equation

(m_1c_1 + m_2c_2)(t_2 - t_0) + m_2'l = m_3c_3(t_0 - t_3)\,. \qquad (10)

However, only the solution satisfying the condition $0 \le m_2' \le m_2$ does make physical sense.

b) Substituting the particular values of $m_1$ , $m_2$ , $m_3$ , $t_2$ and $t_3$ to equations (7), (8) and (9) one obtains solutions not making the physical sense (not satisfying the above conditions for $t$ , respectively $m_3'$ ). The real physical process under given conditions is given by the equation (10) which yields

m_2' = \frac{m_3c_3(t_0 - t_3) - (m_1c_1 + m_2c_2)(t_2 - t_0)}{l}\,.

Substituting given numerical values one gets $m_2' = 0.11\ \mathrm{kg}$ . Hence, $t = 0\,{}^{\circ}\mathrm{C}$ , $M_2 = m_2 - m_2' = 0.89\ \mathrm{kg}$ , $M_3 = m_3 + m_2' = 2.11\ \mathrm{kg}$ .

Оригинал в Архива: ipho69.pdf