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Chapter 12: Problem 5

\(90 \mathrm{gm}\) of ice at \(0^{\circ} \mathrm{C}\) is mixed with \(180 \mathrm{gm}\) of water at \(40^{\circ} \mathrm{C}\). Then the temperature of the resulting mixture is (latent heat of fusion of water is \(80 \mathrm{cal} / \mathrm{gm}\) ) (a) \(0^{\circ} \mathrm{C}\) (b) \(10^{\circ} \mathrm{C}\). (c) \(20^{\circ} \mathrm{C}\). (d) \(30^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The resulting mixture's temperature is \(0^{\circ} \mathrm{C}\).

Step by step solution

01

Calculate Heat Required to Melt Ice

To melt the ice, we need to use the formula for latent heat: \[ Q = m \cdot L_f \] where \( Q \) is the heat required to melt the ice, \( m \) is the mass of the ice, and \( L_f \) is the latent heat of fusion. For this problem: \( m = 90 \; \text{gm} \) and \( L_f = 80 \; \text{cal/gm} \). So, the heat required to melt the ice is: \[ Q = 90 \; \text{gm} \times 80 \; \text{cal/gm} = 7200 \; \text{cal} \]
02

Calculate Heat Lost by Water

The water will lose heat to allow the ice to melt and increase its temperature. For the water cooling from \( 40^{\circ} \text{C} \) to \( 0^{\circ} \text{C} \), we use the formula: \[ Q = m \cdot c \cdot \Delta T \] where \( Q \) is the heat lost, \( m = 180 \; \text{gm} \) is the mass of water, \( c = 1 \; \text{cal/gm}^{\circ} \text{C} \) is the specific heat capacity, and \( \Delta T = 40^{\circ}C \) is the temperature change. Therefore, the heat lost by water is: \[ Q = 180 \; \text{gm} \times 1 \; \text{cal/gm}^{\circ} \text{C} \times 40^{\circ} \text{C} = 7200 \; \text{cal} \]
03

Analyze Heat Exchange

The heat lost by the water (7200 cal) is exactly equal to the heat needed to melt the ice (7200 cal), meaning the ice can completely melt by absorbing all the heat lost from the water, and the resulting mixture will equilibrate at \( 0^{\circ} \text{C} \), since no additional heat is available for raising the temperature further. Thus, the final equilibrium condition is achieved when the entire ice is melted and the temperature stays constant at the initial melting point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Specific Heat Capacity
Specific heat capacity is a crucial property of substances that describes how much heat is required to change the temperature of a particular mass of a substance by one degree Celsius. This concept helps us understand why some materials heat up or cool down faster than others.
  • For water, the specific heat capacity is relatively high, at approximately 1 cal/gm掳C.
  • This means it takes 1 calorie to raise 1 gram of water by 1掳C.
This property is significant in our exercise, where 180 grams of water cool down, losing heat to the surrounding ice. The specific heat capacity reveals how much energy is involved in cooling the water from 40掳C to 0掳C. This calculation is crucial in understanding how heat is transferred in such systems.
Exploring Heat Exchange
Heat exchange is the process where thermal energy is transferred from a hotter substance to a cooler one. In the exercise, the ice and the water are the two substances between which heat exchange occurs.
  • The water, being initially warmer, loses heat.
  • The ice, being colder, absorbs the heat.
These exchanges are pivotal in ensuring that the temperatures of both substances eventually equalize. The fundamental principle here is the conservation of energy. The energy lost by the water equals the energy gained by the ice. Consequently, the final temperature becomes 0掳C as all heat from the water goes into melting the ice.
Delving into Phase Change
Phase change, in this context, refers to the transition of ice from a solid state to a liquid state. This transformation is neither instantaneous nor spontaneous; it requires energy in the form of heat, known as latent heat of fusion.
  • For water, the latent heat of fusion is 80 cal/gm.
  • This is the energy necessary to change 1 gram of ice at 0掳C to 1 gram of water at 0掳C without changing temperature.
In the provided exercise, the ice absorbs a specific amount of heat, equal to the heat lost by the water, to complete its phase change. This process results in the system stabilizing at a temperature where the phase change completes without any excess energy to further increase the temperature. Understanding phase change is vital because it explains why substances remain at a stable temperature while transitioning between physical states.

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Most popular questions from this chapter

A copper cube of mass 200 g slides down a rough inclined plane of inclination \(37^{\circ}\) at a constant speed. Assuming that the loss in mechanical energy goes into the copper block as thermal energy. Find the increas \(\mathrm{c}\) in temperaturc \(o[\) the block as it slides down through \(60 \mathrm{~cm}\). Spccific heat capacity of copper is cqual to \(420 \mathrm{~J} / \mathrm{kg}-\mathrm{K}\). (take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) )

The rales of fall of lemperature of two solid sphercs of same radii bul of diflrent malerials are cqual at certain temperature. (a) Their hoat capacities are cqual. (b) Their heat capacities are proportional to their density. (c) Their specific heat capacities are proportional to their densities. (d) Their specific heal capacities are inversely proportional to their densitics.

\(\Lambda\) pond of water at \(0^{\circ} \mathrm{C}\) is covered with layer of ice \(4 \mathrm{~cm}\) thick if air temperature is \(-10^{\circ} \mathrm{C}\) (constant), how long it takes for ice thickness to increase to \(8 \mathrm{~cm} ? K_{\text {ia }}=2 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}, L_{f}=80 \mathrm{cal} / \mathrm{gm}\), \(\rho_{\text {ice }}=900 \mathrm{~kg} / \mathrm{m}^{3}\)

By which of the following methods could a cup of hot tea lose heat, if placed on metallic table in a classroom? (a) Conduction (b) Convection (c) Radiation (d) Evaporaton of liquid

The temperature of a room is maintained uniform through out the room at \(15^{\circ} \mathrm{C}\) by a heater of resistance 25 ohms connected to 220 volts mains. Calculate the temperature outside if heat transmitted through a glass window of thickness \(0.4 \mathrm{~cm}\). and area 1 sq. metre |Thermal conductivity of glass \(=0.002\) CGS units and \(\mathrm{J}=4.2 \mathrm{~J} / \mathrm{cal} .1\)
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