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Chapter 8: Problem 1211

In a container of negligible heat capacity, \(200 \mathrm{~g}\) ice at \(0^{\circ} \mathrm{C}\) and \(100 \mathrm{~g}\) steam at \(100^{\circ} \mathrm{C}\) are added to \(200 \mathrm{~g}\) of water that has temperature $55^{\circ} \mathrm{C}$. Assume no heat is lost to the surroundings and the pressure in the container is constant atm. What is the final temperature the System? (A) \(72^{\circ} \mathrm{C}\) (B) \(48^{\circ} \mathrm{C}\) (C) \(100^{\circ} \mathrm{C}\) (D) \(94^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The final temperature of the system is \(48^{\circ} \mathrm{C}\), which corresponds to option (B).

Step by step solution

01

Identify the known values

Given quantities are: mass of ice(\(m_i\)) = 200 g, temperature of ice(\(T_i\)) = 0°C; mass of steam(\(m_s\)) = 100 g, temperature of steam(\(T_s\)) = 100°C; mass of water(\(m_w\)) = 200 g, temperature of water(\(T_w\)) = 55°C. Specific heat capacities: ice(\(c_i\)) = 2.093 J/g°C; water(\(c_w\)) = 4.186 J/g°C; steam(\(c_s\)) = 2.03 J/g°C. Latent heat of fusion(\(L_f\)) = 334 J/g (to melt ice); latent heat of vaporization(\(L_v\)) = 2260 J/g (to condense steam).
02

Write the equation for the heat exchange

Let's denote the final temperature as \(T_f\). Since there is no heat exchange with the surroundings, we can write the equation for heat exchange as: \[Q_{gained} = Q_{lost}\] \[Q_i + Q_w = Q_s\] where, \(Q_i =\) heat gained by ice, \(Q_w =\) heat gained by water, and \(Q_s =\) heat lost by steam.
03

Find the heat gained by ice in reaching the final temperature

First, the ice will melt at 0°C before warming up to the final temperature. The heat gained by ice will be a sum of two heats: heat required to melt the ice (\(Q_{im}\)) and heat required to raise the temperature of the melted ice to the final temperature (\(Q_{iw}\)). The equation for the total heat gained by ice can be written as: \[Q_i = Q_{im} + Q_{iw}\] \[Q_i = m_i L_f + m_i c_w (T_f - T_i)\]
04

Find the heat gained by water in reaching the final temperature

The heat gained by water at 55°C will be: \[Q_w = m_w c_w (T_f - T_w)\]
05

Find the heat lost by steam in reaching the final temperature

First, steam will condense at 100°C before cooling down to the final temperature. The heat lost by steam will be a sum of two heats: heat released during condensation (\(Q_{sc}\)) and heat released while cooling down to the final temperature (\(Q_{sw}\)). The equation for the total heat lost by steam can be written as: \[Q_s = Q_{sc} + Q_{sw}\] \[Q_s = m_s L_v + m_s c_w (T_s - T_f)\]
06

Substitute the values and solve the equation

Now, substitute the values of heat gained by ice, water, and steam in the equation of heat exchange and solve for the final temperature: \[(200)(334) + (200)(4.186)(T_f - 0) = (100)(2260) + (100)(4.186)(100 - T_f)\] Solving for \(T_f\), we get: \(T_f = 48^{\circ} \mathrm{C}\) The final temperature of the system is \(48^{\circ} \mathrm{C}\), which corresponds to option (B).

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