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Chapter 8: Problem 1126

\(200 \mathrm{~g}\) of water is heated from $25^{\circ} \mathrm{C}^{\circ} 45^{\circ} \mathrm{C}$ Ignoring the slight expansion of the water the change in its internal energy is (Specific heat of wafer \(1\left\\{(\right.\) cal \(\left.) /\left(9^{\circ} \mathrm{C}\right)\right\\}\) (A) \(33.4 \mathrm{KJ}\) (B) \(11.33 \mathrm{KJ}\) (C) \(5.57 \mathrm{KJ}\) (D) \(16.7 \mathrm{KJ}\)

Short Answer

Expert verified
The change in the internal energy of the water is approximately (D) \(16.7 \mathrm{KJ}\).

Step by step solution

01

1. Identify the variables and formula

Variables: - Mass of water (m) = \(200 \mathrm{~g}\) - Initial temperature (T1) = \(25^{\circ} \mathrm{C}\) - Final temperature (T2) = \(45^{\circ} \mathrm{C}\) - Specific heat of water (c) = \(1\mathrm{~cal/g^{\circ} C}\) Formula: \(Q = m \cdot c \cdot \Delta T\)
02

2. Convert specific heat to an appropriate unit

We will calculate the internal energy in \(\mathrm{KJ}\), so let's convert the specific heat of water from \(\mathrm{cal/g^{\circ} C}\) to \(\mathrm{J/g^{\circ} C}\) using the conversion factor \(1\mathrm{~cal} = 4.184 \mathrm{~J}\). c (in J) = \(1\mathrm{~cal/g^{\circ} C} \times 4.184 \mathrm{~J/cal}\) = \(4.184 \mathrm{~J/g^{\circ} C}\)
03

3. Calculate the temperature difference (\(\Delta T\))

\(\Delta T\) is the difference between the final and initial temperatures. \(\Delta T = T2 - T1 = 45^{\circ} \mathrm{C} - 25^{\circ} \mathrm{C} = 20^{\circ} \mathrm{C}\)
04

4. Calculate the change in internal energy (Q)

Now, we can use the formula with the variables and temperature difference to find the change in internal energy. \(Q = m \cdot c \cdot \Delta T = 200\mathrm{~g} \times 4.184 \mathrm{~J/g^{\circ} C} \times 20^{\circ} \mathrm{C} = 16736 \mathrm{~J}\)
05

5. Convert the result to KJ and choose the correct option

Finally, we convert the obtained value to \(\mathrm{KJ}\): \(Q = \frac{16736 \mathrm{~J}}{1000} = 16.736 \mathrm{~KJ}\) Thus, the change in the internal energy of the water is approximately (D) \(16.7 \mathrm{KJ}\).

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