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Chapter 15: Problem 5

\(250 \mathrm{gm}\) of water and equal volume of alcohol of mass \(200 \mathrm{gm}\) are replaced successively in the same calorimeter and cool from \(60^{\circ} \mathrm{C}\) to \(55^{\circ} \mathrm{C}\) in 130 seconds and 67 seconds respectively. If the water equivalent of the calorimeter is \(10 \mathrm{gm}\), then the specific heat of alcohol (in \(\mathrm{cal} / \mathrm{gm}-{ }^{\circ} \mathrm{C}\) ) is (a) \(1.30\) (b) \(0.67\) (c) \(0.62\) (d) \(0.985\)

Short Answer

Expert verified
The specific heat of alcohol is approximately 0.62. Option (c) is correct.

Step by step solution

01

Determine Heat Lost

Calculate the heat lost by water as it cools from \(60^{\circ} \mathrm{C}\) to \(55^{\circ} \mathrm{C}\). The formula for heat loss is given by:\[ Q = (m_w + W) \cdot c_w \cdot \Delta T \]where - \(m_w\) is the mass of water, - \(W\) is the water equivalent of the calorimeter, - \(c_w\) is the specific heat of water (\(1 \ \mathrm{cal} / \mathrm{gm}^{\circ} \mathrm{C}\)),- \(\Delta T\) is the change in temperature.Using the values,\[ Q = (250 + 10) \cdot 1 \cdot (60 - 55) = 260 \times 5 = 1300 \ \mathrm{cal} \].
02

Calculate Rate of Heat Loss for Water

The rate of heat loss for water can be calculated by:\[ Q_w = \frac{1300 \ \mathrm{cal}}{130 \ \mathrm{seconds}} = 10 \ \mathrm{cal/sec} \]
03

Determine Heat Lost for Alcohol

Assuming the rate of heat loss for alcohol and water are the same due to identical conditions of the calorimeter, we use:\[ Q_a = (m_a + W) \cdot c_a \cdot \Delta T = 10 \cdot 67 = 670 \ \mathrm{cal} \] ,where - \(m_a\) is the mass of the alcohol,- \(c_a\) is the specific heat of alcohol. We know the heat loss \(Q_a = 670\).
04

Solve for Specific Heat of Alcohol

Given \[ 670 = (200 + 10) \cdot c_a \cdot (60 - 55) \].Simplify and solve for \(c_a\):\[ 670 = 210c_a \cdot 5 \] \[ 670 = 1050c_a \]\[ c_a = \frac{670}{1050} \approx 0.638 \ \mathrm{cal} / \mathrm{gm}^{\circ} \mathrm{C} \].Upon comparison with the given options, the closest match is option \(c) 0.62\).

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Key Concepts

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

Heat Loss Calculation
When substances change temperature, they lose or gain heat. Understanding heat loss is crucial. For our problem, as water cools, it loses heat. The formula for heat loss is \[ Q = (m + W) \cdot c \cdot \Delta T \]. This formula helps calculate the heat lost when a substance cools. Let's break down the involved terms:
  • \(m\) is the mass of the substance.
  • \(W\) is the water equivalent, a factor representing how much the calorimeter can absorb.
  • \(c\) is the specific heat capacity, determining how much heat is needed to change a unit mass by one-degree Celsius.
  • \(\Delta T\) is the change in temperature.
In our example, this calculation for water is straightforward, resulting in 1300 calories of heat loss.
Calorimetry
Calorimetry is the science of measuring heat. It relies on understanding how heat flows between objects. By conducting experiments with water and alcohol, we can determine important properties of materials, like specific heat capacity. In our scenario, a calorimeter finds the heat flow using the temperature change.
  • We measure the substance's temperature before and after heat exchange.
  • The calorimeter's role is pivotal. It ensures consistent conditions for those measurements.
Through these steps, insights like specific heat capacity become clear, helping identify material properties.
Specific Heat Capacity
Each material has a unique specific heat capacity. This property tells us how much heat energy is needed to raise its temperature. A high specific heat means the substance can absorb a lot of heat without much temperature change. In our context:
  • The specific heat of water is known as 1 \(\mathrm{cal/gm}^{\circ} \mathrm{C}\).
  • Alcohol, being different, needs its specific heat calculated.
By using the heat loss formula and considering the mass and temperature change, the specific heat capacity of alcohol is computed to be approximately 0.62 \( \mathrm{cal/gm}^{\circ} \mathrm{C}\). This property helps in understanding how alcohol behaves thermally.
Cooling Time
Cooling time is how long it takes for a substance to change temperature under cooling. This time is crucial in experiments to compare material responses.In this experiment:
  • Water cools from \(60^{\circ} \mathrm{C} \) to \(55^{\circ} \mathrm{C}\) in 130 seconds.
  • Alcohol cools the same amount in 67 seconds.
This different cooling time reflects their thermal properties. For example, alcohol cools faster due to its lower specific heat capacity. Knowing these cooling times is beneficial to control processes involving heat exchange.

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Most popular questions from this chapter

Three objects coloured black, grey and white can withstand hostile conditions upto \(2800^{\circ} \mathrm{C}\). These objects are thrown into a furnace where each of them attains a temperature of \(2000^{\circ} \mathrm{C}\). Which object will glow the brightest? (a) The white object (b) The black object (c) All glow with equal brightness (d) Grey object

An electrically heated coil is immersed in a calorimeter containing \(360 \mathrm{gm}\) of water at \(10^{\circ} \mathrm{C}\). The coil consumes energy at the rate of 90 watts. The water equivalent of calorimeter and coil is \(40 \mathrm{gm}\). The temperature of water after 10 minutes is (a) \(4.214^{\circ} \mathrm{C}\) (b) \(42.14^{\circ} \mathrm{C}\) (c) \(30^{\circ}\) (d) none of these

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