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How many grams of ice at \(-14{ }^{\circ} \mathrm{C}\) are needed to cool \(200 \mathrm{~g}\) of water from \(25^{\circ} \mathrm{C}\) to \(10^{\circ} \mathrm{C} ?\) Take spccific heat of icc \(=0.5 \mathrm{cal} / \mathrm{g} /{ }^{\circ} \mathrm{C}\) and latent heat o \(\mathrm{icc}=80 \mathrm{cal} / \mathrm{g}\).

Short Answer

Expert verified
34.48 grams of ice are needed.

Step by step solution

01

Determine Heat Removed by Water

To find the total heat that must be removed from the water, calculate it using the formula for specific heat: \(Q = mc\Delta T\). Here, \(m = 200\, \text{g}\), \(c = 1\, \text{cal/g/} ^{\circ} \text{C}\), and the temperature change \(\Delta T = (10 - 25) ^{\circ} \text{C} = -15 \, ^{\circ} \text{C}\). Thus, \(Q = 200 \times 1 \times (-15) = -3000\, \text{cal}\). The negative sign indicates heat is lost.
02

Calculate Heat Gained by Ice to Warm to 0°C

The ice at \(-14^{\circ} \text{C}\) needs to warm up to \(0^{\circ} \text{C}\). Use the formula \(Q = mc\Delta T\) with \(c = 0.5 \text{ cal/g/} ^{\circ} \text{C}\) and \(\Delta T = 0 - (-14)\, ^{\circ} \text{C} = 14\, ^{\circ} \text{C}\). The heat required is \(Q = m \times 0.5 \times 14 = 7m\, \text{cal}\).
03

Calculate Heat Gained by Ice to Melt

When the ice has warmed up to \(0^{\circ} \text{C}\), it needs additional heat to melt. The heat required to convert ice to water at \(0^{\circ} \text{C}\) is given by \(Q = mL\). Here, \(L = 80\, \text{cal/g}\), so \(Q = m \times 80\).
04

Set Up Heat Balance Equation

The heat lost by the water must equal the total heat gained by the ice (to heat up and melt it). These energies are: \[3000 = 7m + 80m\] Simplifying gives: \[3000 = 87m\].
05

Solve for Ice Mass

Solve the equation \(3000 = 87m\) for \(m\) by dividing both sides by 87: \[m = \frac{3000}{87} \approx 34.48\, \text{g}\].

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Key Concepts

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

Understanding Latent Heat
Latent heat is an incredible concept in thermodynamics, often described as the heat required to change the state of a substance without changing its temperature. For example, when ice melts to become water, it absorbs heat but remains at the same temperature. This is latent heat in action.

When ice melts at its melting point, it requires a specific amount of energy known as the latent heat of fusion. For ice, this value is about 80 cal/g. This means that 80 calories of energy are needed for each gram of ice to transform it into water while the temperature holds steady at 0°C.

In our problem, as the ice warms up and reaches 0°C, it takes additional energy to transition into a liquid state. Calculating this requires multiplying the mass of the ice by the latent heat constant. Latent heat is a key player in processes where heat energy doesn’t affect temperature but facilitates phase transition.
How Heat Transfer Works
Heat transfer is the mechanism of energy moving from a warmer object to a cooler one. In our daily lives, this might be something as simple as an ice cube melting in warm water.

The basic principle is that heat naturally flows from objects of higher temperature to those of lower temperature in order to reach equilibrium. In our exercise, the water, originally at 25°C, will discharge some of its thermal energy to the colder ice to match temperatures.

The process can be calculated through the formula \[ Q = mc\Delta T \]. This formula helps gauge the heat exchange based on specific heat capacity (\( c \)). This principle ensures that energy balances out, meaning no net change in total energy. Understanding how much energy is shifted is essential in thermodynamics and helps in predicting changes in temperature during various physical processes.
Reaching Thermal Equilibrium
Thermal equilibrium occurs when two substances reach the same temperature and no further heat transfer takes place. It's like having a balance on a seesaw with everything perfectly aligned.

In our problem, the goal is to achieve thermal equilibrium between the ice and water. Initially, the water is warmer and the ice cooler, so heat flows from the water to the ice. As the ice absorbs heat, it melts and may eventually become the same temperature as the water. Once this balance is struck, we've reached thermal equilibrium.

Understanding this concept is essential for tasks like predicting how long your drink will stay cold, or in industrial processes like cooling systems. The concept ties directly to energy conservation, as reaching equilibrium involves a mindful check of energy transfer rates between substances.

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Most popular questions from this chapter

Therc are four vernicr scalcs; whosc specifications are given in column-I and the least count is given in Column-II \((S=\) valuc of main scale division, \(n=\) number of marks on vemicr) \begin{tabular}{l|l} Column-I & Column-II \end{tabular} (a) \(S=1 \mathrm{~mm}, n=10\) (p) \(0.05 \mathrm{~mm}\) (b) \(S=0.5 \mathrm{~mm}, n=10\) (q) \(0.01 \mathrm{~mm}\) (c) \(S=0.5 \mathrm{~mm}, n=20\) (r) \(0.1 \mathrm{~mm}\) (d) \(S=1 \mathrm{~mm}, n=100\) (s) \(0.025 \mathrm{~mm}\)

The temperaturc of cqual masses of three differen liquids \(\Lambda, B\) and \(C\) are \(12^{\circ} \mathrm{C}\), \(19^{\circ} \mathrm{C}\) and \(28^{\circ} \mathrm{C}\). respectively. The temperature when \(A\) and \(B\) are mixed is \(16^{\circ} \mathrm{C}\) and when \(B\) and \(C\) are mixed it is \(23^{\circ} \mathrm{C}\). What should be the temperature when \(\Lambda\) and \(C\) are mixed?

Two ends of a conducting rod of varying cross-section are maintained at \(200^{\circ} \mathrm{C}\) and \(0^{\circ} \mathrm{C}\) respectively. In steady state (a) temperature difference across \(A B\) and \(C D\) are equal (b) temperature difference across \(A B\) is greater than that of across \(C D\) (c) temperature difference across \(A B\) is less than that of across \(C D\) (d) temperature difference may be equal or different depending on the thermal conductivity of the rod

The temperature drop through a two layer furnace wall is \(900^{\circ} \mathrm{C}\). Lach laycr is of cqual arca of cross-section. Which of the following actions will result in lowering the temperature \(\theta\) of the interface? (a) By increasing the thermal conductivity of ouler laycr (b) By increasing the thermal conductivity of inner (c) By increasing thickness of outer layer (d) By increasing thickness of inner layer

Which of the following statements are true? (a) A material of high conductivity can absorb large amount of heat. (b) The coefficient ol thermal conductivity cannot determine the dircetion of propagation ol heat. (c) A good conductor loses heat faster, (d) Free electrons transport heat from higher temperature to lower temperature,

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