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Chapter 30: Problem 2840

\(100 \mathrm{~g}\) ice at \(0^{\circ} \mathrm{C}\) placed in \(100 \mathrm{~g}\) water at \(100^{\circ} \mathrm{C}\). The final temperature of the mixture will be........ (Latent heat of ice is \(80 \mathrm{Cal} / \mathrm{g}\), and specific heat of water is \(1 \mathrm{Cal} / \mathrm{g} \mathrm{C}^{\circ}\) ) (A) \(10^{\circ} \mathrm{C}\) (B) \(20^{\circ} \mathrm{C}\) (C) \(30^{\circ} \mathrm{C}\) (D) \(50^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The final temperature of the mixture is (B) \(20^{\circ} \mathrm{C}\).

Step by step solution

01

Write down the heat gained by ice and heat lost by water

To calculate the final temperature, we need to find the heat gained by ice when it turns into water and the heat lost by water when it cools down in the process. The heat transfer process can be described with the following equation: Heat gained by ice = Heat lost by water
02

Express heat gained by ice in terms of mass, latent heat, and specific heat of water

When the ice turns into water at \(0^{\circ} \mathrm{C}\), it will gain heat due to its latent heat and then gain more heat to reach the final temperature. So, the heat equation of ice becomes: \( Q_{ice} = m_{ice} \times L + m_{ice} \times c \times (T_f - 0) \) Here, \(m_{ice}\) - mass of ice (\(100 \mathrm{g}\)) \(\ L\) - latent heat of ice (\(80 \mathrm{Cal} / \mathrm{g}\)) \(c\) - specific heat of water (\(1 \mathrm{Cal} / \mathrm{g} \mathrm{C}^{\circ}\)) \(T_f\) - final temperature
03

Express heat lost by water in terms of mass, specific heat, and final temperature

The water's heat equation when cooling down from \(100^{\circ} \mathrm{C}\) to the final temperature is: \( Q_{water} = m_{water} \times c \times (100 - T_f) \) Here, \(m_{water}\) - mass of water (\(100 \mathrm{g}\))
04

Set the heat gained by ice equal to the heat lost by water and solve for the final temperature

Using the principle of conservation of energy, set the heat gained by ice equation equal to the heat lost by water equation and solve for the final temperature: \( m_{ice} \times L + m_{ice} \times c \times (T_f - 0) = m_{water} \times c \times (100 - T_f) \) Now, plug in the given values: \( 100 \times 80 + 100 \times 1 \times (T_f - 0) = 100 \times 1 \times (100 - T_f) \) By solving the above equation, we get the final temperature: \( T_f = 20^{\circ} \mathrm{C} \) So, the answer is (B) \(20^{\circ} \mathrm{C}\).

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Most popular questions from this chapter

Amount of heat required to raise the temperature of a body through $1 \mathrm{k}$ is called its (A) water equivalent (B) Thermal capacity (C) entropy (D) specific heat

The temperature of equal masses of three different liquids \(\mathrm{X}\), \(\mathrm{Y}, \mathrm{Z}\) are \(12^{\circ} \mathrm{C}, 19^{\circ} \mathrm{C}\) and \(28^{\circ} \mathrm{C}\) respectively. The temperature when \(\mathrm{X}\) and \(\mathrm{Y}\) are mixed is \(16^{\circ} \mathrm{C}\) and when \(\mathrm{Y}\) and \(\mathrm{Z}\) are mixed is \(23^{\circ} \mathrm{C}\) what is the temperature when \(\mathrm{X}\) and \(\mathrm{Z}\) are mixed ? (A) \(21.6^{\circ} \mathrm{C}\) (B) \(18.5^{\circ} \mathrm{C}\) (C) \(23.25^{\circ} \mathrm{C}\) (D) \(20.3^{\circ} \mathrm{C}\)

Two liquids of equal volume are thoroughly mixed. If their specific heat are \(c_{1}, c_{2}\), temperatures \(\theta_{1}, \theta_{2}\) and densities \(\mathrm{d}_{1}, \mathrm{~d}_{2}\) respectively. What is the final temperature of the mixture ? (A) $\left\\{\left(\mathrm{d}_{1} \mathrm{c}_{1} \theta_{1}+\mathrm{d}_{2} \mathrm{c}_{2} \theta_{2}\right) /\left(\mathrm{d}_{1} \theta_{1}+\mathrm{d}_{2} \theta_{2}\right)\right\\}$ (B) $\left\\{\left(\mathrm{c}_{1} \theta_{1}+\mathrm{c}_{2} \theta_{2}\right) /\left(\mathrm{d}_{1} \mathrm{c}_{1}+\mathrm{d}_{2} \mathrm{c}_{2}\right)\right\\}$ (C) $\left\\{\left(\mathrm{d}_{1} \theta_{1}+\mathrm{d}_{2} \theta_{2}\right) /\left(\mathrm{c}_{1} \theta_{1}+\mathrm{c}_{2} \theta_{2}\right)\right\\}$ (D) $\left\\{\left(\mathrm{d}_{1} \mathrm{c}_{1} \theta_{1}+\mathrm{d}_{2} \mathrm{c}_{2} \theta_{2}\right) /\left(\mathrm{d}_{1} \mathrm{c}_{1}+\mathrm{d}_{2} \mathrm{c}_{2}\right)\right\\}$

A centigrade and a Fahrenheit thermometer one dipped in boiling water. The water temperature is lowered until the Fahrenheit thermometer registers \(140^{\circ}\). What is the fall in temperatures as registered by the centigrade thermometer? (A) \(30^{\circ}\) (B) \(40^{\circ}\) (C) \(60^{\circ}\) (D) \(80^{\circ}\)
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