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Chapter 14: Problem 19

A \(0.400-\mathrm{kg}\) aluminum teakettle contains \(2.00 \mathrm{kg}\) of water at \(15.0^{\circ} \mathrm{C} .\) How much heat is required to raise the temperature of the water (and kettle) to \(100.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Answer: The total heat energy required to raise the temperature of the water and the aluminum teakettle to 100.0掳C is 739,900 J.

Step by step solution

01

Identify given values and variables

Given values: Mass of aluminum teakettle (m1) = 0.400 kg Mass of water (m2) = 2.00 kg Initial temperature (Ti) = 15.0掳C Final temperature (Tf) = 100.0掳C Specific heat capacity of aluminum (c1) = 900 J/kg掳C Specific heat capacity of water (c2) = 4186 J/kg掳C Variables to find: Heat energy required to raise the temperature of the water (q1) Heat energy required to raise the temperature of the aluminum teakettle (q2) Total heat energy required (Q)
02

Find the change in temperature

螖T = Tf - Ti = 100.0 - 15.0 = 85.0掳C
03

Calculate the heat energy required to raise the temperature of the water

Using the formula \(q = mc螖T\), we can find the heat energy required for the water. q1 = m2 * c2 * 螖T = 2.00 * 4186 * 85.0 = 709300 J
04

Calculate the heat energy required to raise the temperature of the aluminum teakettle

Similarly, we can find the heat energy required for the aluminum teakettle. q2 = m1 * c1 * 螖T = 0.400 * 900 * 85.0 = 30600 J
05

Find the total heat energy required

To find the total heat energy required, we add the heat energy required for the water and the aluminum teakettle. Q = q1 + q2 = 709300 J + 30600 J = 739900 J So, the total heat energy required to raise the temperature of the water (and the aluminum teakettle) to 100.0掳C is 739900 J.

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