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Chapter 2: Problem 126

How many joules of heat energy would it take to raise the temperature of \(2.00\) pounds of iron from \(30.0^{\circ} \mathrm{C}\) to \(90.0^{\circ} \mathrm{C} ?[453.6 \mathrm{~g}=1\) pound \(]\)

Short Answer

Expert verified
The heat energy required to raise the temperature of $2.00$ pounds of iron from $30.0^{\circ} \mathrm{C}$ to $90.0^{\circ} \mathrm{C}$ is \(24,393.6 \, \mathrm{J}\).

Step by step solution

01

Convert the mass from pounds to kilograms

First, we need to convert the mass of the iron from pounds to kilograms. We are given the conversion factor: 1 pound is equal to 453.6 grams, and 1 kilogram is equal to 1000 grams. So, \(2.00 \, lb \times \frac{453.6 \, g}{1 \, lb} \times \frac{1 \, kg}{1000 \, g} = 0.9072 \, kg\) The mass of the iron is 0.9072 kg.
02

Calculate the change in temperature

The change in temperature (\(\Delta T\)) can be found by subtracting the initial temperature from the final temperature: \(\Delta T = T_{final} - T_{initial} = 90.0^\circ C - 30.0^\circ C = 60.0^\circ C\) The change in temperature is 60.0 °C.
03

Calculate the heat energy

Now, we can use the heat energy formula, along with the mass, specific heat capacity, and change in temperature: \(Q = mc\Delta T\) \(Q = (0.9072 \, kg) (450 \, \frac{J}{kg \cdot^\circ C}) (60.0 \, ^\circ C)\) \(Q = 24393.6 \, J\) The heat energy required to raise the temperature of the iron is 24,393.6 joules.

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