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Heat required to raise the temperature of \(1 \mathrm{~mol}\) of a substance by \(1^{\circ}\) is called (a) specific heat (b) molar heat capacity (c) water equivalent (d) specific gravity

Short Answer

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Molar heat capacity (b)

Step by step solution

01

Understand the Terminology

Let's first understand what each term in the options means. - **Specific Heat**: This is the amount of heat required to raise the temperature of 1 gram of a substance by 1°C. - **Molar Heat Capacity**: This is the amount of heat required to raise the temperature of 1 mole of a substance by 1°C. - **Water Equivalent**: This is the amount of water that would absorb the same amount of heat as the substance when their temperatures are changed equally. - **Specific Gravity**: This is the ratio of the density of a substance to the density of a reference substance (usually water for liquids).
02

Identify the Correct Term Based on the Definition

The exercise asks for the heat required to raise the temperature of 1 mole of a substance by 1°C. This definition directly matches **Molar Heat Capacity**.
03

Conclusion

Based on the definition provided in the problem and the meaning of the terms, the correct answer is (b) molar heat capacity. This is because this term specifically refers to the heat required to raise the temperature of one mole of a substance by one degree Celsius.

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Key Concepts

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

Specific Heat
Specific heat is a measure of how much heat energy is needed to change the temperature of a specific substance. It is defined as the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius.
The specific heat capacity can vary between different substances. This is why water, for example, has a very high specific heat capacity compared to metals. Metals typically heat up and cool down faster than water.
  • Specific heat is denoted by the symbol "c" and is measured in units of joules per gram per degree Celsius (J/g°C).
  • The formula to calculate the heat ( \( q \)) absorbed or released is: \( q = mc\Delta T \), where \( m \) is the mass, \( c \) is the specific heat, and \( \Delta T \) is the change in temperature.
By understanding specific heat, students can comprehend how substances like water can moderate temperature, leading to applications like temperature regulation in environments.
Water Equivalent
Water equivalent is a concept that relates to the capacity of a substance to absorb heat compared to that of water. It is essentially the mass of water that would absorb the same amount of heat as the actual substance when both undergo a temperature change.
This concept is particularly useful in calorimetry, where the heat capacity of various materials needs to be compared using a common reference (water).
  • Water equivalent is calculated using the formula: Water Equivalent = mass of substance x specific heat of substance / specific heat of water.
  • This allows for a clear comparison of the thermal capacity of different materials relative to water.
It simplifies the comparison in energy-transfer applications where understanding how materials behave thermally becomes essential.
Specific Gravity
Specific gravity is a ratio that describes the density of a substance in comparison to the density of a reference material, typically water in the case of liquids.
It's a dimensionless quantity (it has no units) as it is the ratio of two similar kinds of measurements.
Specific gravity is a critical parameter in various practical fields such as material science and chemistry, particularly when comparing materials for buoyancy or concentrating solutions.
  • The specific gravity is calculated using the formula: Specific Gravity = Density of substance / Density of water (at 4°C).
  • A specific gravity greater than 1 indicates that the substance is denser than water.
  • If it is less than 1, the substance is less dense and will float when placed in water.
Understanding specific gravity helps students evaluate how substances are likely to behave when mixed or when assessing buoyancy in fluids.

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Most popular questions from this chapter

The enthalpy changes for the following processes are listed below. \(\mathrm{Cl}_{2}(\mathrm{~g})=2 \mathrm{C} 1(\mathrm{~g}) ; 242.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\mathrm{I}_{2}(\mathrm{~g})=21(\mathrm{~g}) ; 151.0 \mathrm{kJmol}^{-1}\) \(\mathrm{ICl}(\mathrm{g})=\mathrm{I}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}) ; 211.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\mathrm{I}_{2}(\mathrm{~s})=\mathrm{I}_{2}(\mathrm{~g}) ; 62.76 \mathrm{~kJ} \mathrm{~mol}^{-1}\) Given that the standard states for iodine and chlorine are \(\mathrm{I}_{2}\) (s) and \(\mathrm{Cl},(\mathrm{g})\), the standard enthalpy of formation for \(\mathrm{ICl}(\mathrm{g})\) is [2006] (a) \(-14.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-16.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(+16.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(+244.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The enthalpy change involved in oxidation of glucose is \(-2880 \mathrm{~kJ} \mathrm{~mol}^{-1}, 25 \%\) of this energy is available for muscular work. If \(100 \mathrm{~kJ}\) of muscular work is needed to walk one \(\mathrm{km}\), the maximum distance \((\mathrm{km})\) that a person will be able to walk after taking \(150 \mathrm{~g}\) of glucose is

Calculate \(\mathrm{Q}\) and \(\mathrm{W}\) for the isothermal reversible expansion of one mole of an ideal gas from an initial pressure of \(1.0\) bar to a final pressure of \(0.1\) bar at a constant temperature of \(273 \mathrm{~K}\). (a) \(5.22 \mathrm{~kJ},-5.22 \mathrm{~kJ}\) (b) \(-27.3 \mathrm{~kJ}, 27.3 \mathrm{~kJ}\) (c) \(27.3 \mathrm{~kJ},-27.3 \mathrm{~kJ}\) (d) \(-5.22 \mathrm{~kJ}, 5.22 \mathrm{~kJ}\)

In monoatomic gases, ratio of specific heat at constant pressure to that at constant volume is (a) \(3 / 5\) (b) \(5 / 3\) (c) \(7 / 5\) (d) \(4 / 5\)

The increase in internal energy of the system is 100 \(\mathrm{J}\) when \(300 \mathrm{~J}\) of heat is supplied to it. What is the amount of work done by the system (a) \(-200 \mathrm{~J}\) (b) \(+200 \mathrm{~J}\) (c) \(-300 \mathrm{~J}\) (d) \(-400 \mathrm{~J}\)

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