Question

The height of victoria falls is 63 m. What is the difference in temperature of water at the top and at the bottom of fall?
[Given 1 cal = 4.2 J and specific heat of water = 1 cal g\(^{-1}\) \(^{\circ}\)C\(^{-1}\)]

• A. 0.014\(^{\circ}\)C
• B. 0.147\(^{\circ}\)C
• C. 1.476\(^{\circ}\)C
• D. 14.76\(^{\circ}\)C

Hint:

For this type of problem, the formula \( \Delta T = \frac{gh}{c} \) is very useful. It is important to ensure all quantities are in SI units before calculating. A quick check of units: \( \frac{(\text{m/s}^2)(\text{m})}{\text{J/kg}\cdot^{\circ}\text{C}} = \frac{\text{m}^2/\text{s}^2}{(\text{kg}\cdot\text{m}^2/\text{s}^2)/\text{kg}\cdot^{\circ}\text{C}} = \frac{\text{m}^2/\text{s}^2}{\text{m}^2/\text{s}^2\cdot^{\circ}\text{C}} = ^{\circ}\text{C} \). The units are consistent.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are asked to find the temperature increase of water after falling from a certain height. This is a problem of energy conservation, where the potential energy of the water at the top is converted into heat energy at the bottom, causing a temperature rise.
Step 2: Key Formula or Approach:
1. Potential Energy (PE): For a mass \(m\) at height \(h\), \( PE = mgh \).
2. Heat Energy (Q): To raise the temperature of mass \(m\) by \( \Delta T \), the heat required is \( Q = mc\Delta T \), where \(c\) is the specific heat capacity.
3. Conservation of Energy: We assume that all the potential energy is converted into heat energy. So, \( PE = Q \).
Step 3: Detailed Explanation:
Given values:
Height, \( h = 63 \, \text{m} \).
Acceleration due to gravity, \( g \approx 9.8 \, \text{m/s}^2 \).
Specific heat of water, \( c = 1 \, \text{cal g}^{-1} \,^{\circ}\text{C}^{-1} \).
First, we need to convert the specific heat capacity to SI units (J kg\(^{-1}\) K\(^{-1}\) or J kg\(^{-1}\) \(^{\circ}\text{C}^{-1}\)).
\( c = 1 \frac{\text{cal}}{\text{g} \cdot ^{\circ}\text{C}} = 1 \frac{4.2 \, \text{J}}{10^{-3} \, \text{kg} \cdot ^{\circ}\text{C}} = 4200 \, \text{J kg}^{-1} \,^{\circ}\text{C}^{-1} \).
Now, we apply the principle of energy conservation. Let's consider a mass \(m\) of water falling.
Loss in Potential Energy = \( mgh \)
Gain in Heat Energy = \( mc\Delta T \)
Equating the two:
\[ mgh = mc\Delta T \] The mass \(m\) cancels out from both sides:
\[ gh = c\Delta T \] Now, we can solve for the temperature difference \( \Delta T \).
\[ \Delta T = \frac{gh}{c} \] Substituting the values:
\[ \Delta T = \frac{9.8 \, \text{m/s}^2 \times 63 \, \text{m}}{4200 \, \text{J kg}^{-1} \,^{\circ}\text{C}^{-1}} \] \[ \Delta T = \frac{617.4}{4200} \, ^{\circ}\text{C} \] \[ \Delta T \approx 0.147 \, ^{\circ}\text{C} \] Step 4: Final Answer:
The difference in temperature of the water is 0.147\(^{\circ}\)C.