Ch. 19 - Heat and the First Law of Thermodynamics

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 19 - Heat and the First Law of Thermodynamics

Problem 18

Chapter 19, Problem 18

A 215-g sample of a substance is heated to 330°C and then plunged into a 105-g aluminum calorimeter cup containing 185 g of water and a 17-g glass thermometer at 10.5°C. The final temperature is 35.0°C. What is the specific heat of the substance? (Assume no water boils away.)

Verified step by step guidance

Convert all given masses to kilograms (kg) since the SI unit for mass is kilograms. For example, the mass of the substance is 215 g, which is equivalent to 0.215 kg.

Write the heat transfer equation for the system. The heat lost by the substance is equal to the heat gained by the aluminum calorimeter, water, and glass thermometer. Mathematically, this can be expressed as:

Q_{substance} = Q_{aluminum} + Q_{water} + Q_{glass}

Express the heat transfer for each component using the formula

Q = mc\(\Delta\) T

, where

m

is the mass,

c

is the specific heat capacity, and

\(\Delta\) T

is the temperature change. For the substance:

Q_{substance} = m_{substance} c_{substance} (T_{initial,substance} - T_{final})

. For aluminum:

Q_{aluminum} = m_{aluminum} c_{aluminum} (T_{final} - T_{initial,aluminum})

. Similarly, write expressions for water and glass.

Substitute the known values into the equations. Use the specific heat capacities:

c_{aluminum} = 900 \ \(\text{J/kg·°C}\)

c_{water} = 4186 \ \(\text{J/kg·°C}\)

, and

c_{glass} = 840 \ \(\text{J/kg·°C}\)

. The initial and final temperatures are given in the problem.

Solve for the specific heat capacity of the substance,

c_{substance}

, by rearranging the equation:

c_{substance} = \(\frac{Q_{aluminum}\) + Q_{water} + Q_{glass}}{m_{substance} (T_{initial,substance} - T_{final})}

. Perform the algebraic simplifications to isolate

c_{substance}

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Specific Heat Capacity

Specific heat capacity is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. It is a material property that varies between different substances and is crucial for calculating heat transfer in thermal processes. The formula used is Q = mcΔT, where Q is the heat absorbed or released, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Heat Transfer

Heat transfer refers to the movement of thermal energy from one object or substance to another due to a temperature difference. In this scenario, heat is transferred from the heated substance to the cooler water, aluminum cup, and thermometer until thermal equilibrium is reached. Understanding the principles of heat transfer is essential for solving calorimetry problems, as it allows us to apply the conservation of energy.

Calorimetry

Calorimetry is the science of measuring the heat of chemical reactions or physical changes, as well as heat capacity. In this problem, a calorimeter is used to measure the heat exchange between the heated substance and the water, cup, and thermometer. By applying the principle of conservation of energy, where the heat lost by the substance equals the heat gained by the water and calorimeter, we can determine the specific heat of the unknown substance.

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