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Chapter 7: Problem 1069

In a water fall the water falls from a height of \(100 \mathrm{~m}\). If the entire K.E. of water is converted in to heat the rise in temperature of water will be (A) \(0.23^{\circ} \mathrm{C}\) (B) \(0.46^{\circ} \mathrm{C}\) (C) \(2.3^{\circ} \mathrm{C}\) (D) \(0.023^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The rise in temperature of the water when the entire kinetic energy is converted into heat is approximately \(0.23^{\circ} \mathrm{C}\).

Step by step solution

01

Identify the given information

We are given the height of the waterfall (h = 100 m).
02

Calculate the Potential Energy of the Water

We will use the potential energy equation to calculate the potential energy of the water falling from the height given. The potential energy equation is: PE = m * g * h Where: PE = potential energy (Joules) m = mass of the water (kg) g = gravitational acceleration (9.81 m/s²) h = height of the waterfall (100 m) We don't have a value for the mass of the water (m), but since we only need the change in temperature, which is a ratio, the mass will cancel out in our equation.
03

Find the Heat Energy Equation

We know that the kinetic energy (KE) of the water is converted into heat energy. From the law of conservation of energy, the potential energy (PE) will be equal to the heat energy (Q). The heat energy equation is: Q = m * C * ΔT Where: Q = heat energy (Joules) m = mass of the water (kg) C = specific heat capacity of water (4.18 J/g°C, remember that 1 J/g°C = 1 kJ/kg°C, so C = 4.18 kJ/kg°C) ΔT = change in temperature (°C) Since we know that PE = Q, we can relate the equations: m * g * h = m * C * ΔT
04

Calculate the Change in Temperature

Now we will solve for ΔT (change in temperature). First, we can cancel out mass (m) on both sides, which gives us: g * h = C * ΔT Then we can plug in the values for g, h, and C: (9.81 m/s²) * (100 m) = (4.18 kJ/kg°C) * ΔT Now, divide by the specific heat capacity (C) to solve for ΔT: (9.81 * 100) / 4.18 = ΔT Simplify: \( \Delta T = \frac{981}{4.18} = 234.69 / 1000 = 0.23469 °C \)
05

Choose the Correct Option

Now, we can compare our result to the given options: (A) \(0.23^{\circ} \mathrm{C}\) (B) \(0.46^{\circ} \mathrm{C}\) (C) \(2.3^{\circ} \mathrm{C}\) (D) \(0.023^{\circ} \mathrm{C}\) The answer most closely resembles option A: The rise in temperature of the water when the entire kinetic energy is converted into heat is approximately \(0.23^{\circ} \mathrm{C}\).

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Most popular questions from this chapter

Radius of a soap bubble is \(\mathrm{r}^{\prime}\), surface tension of soap solution is \(\mathrm{T}\). Then without increasing the temperature how much energy will be needed to double its radius. (A) \(4 \pi r^{2} T\) (B) \(2 \pi r^{2} T\) (C) \(12 \pi r^{2} T\) (D) \(24 \pi r^{2} T\)

At what temperature the centigrade (celsius) and Fahrenheit readings at the same. \((\mathrm{A})-40^{\circ}\) (B) \(+40^{\circ} \mathrm{C}\) (C) \(36.6^{\circ}\) (D) \(-37^{\circ} \mathrm{C}\)

What is the relationship between Young's modulus Y, Bulk modulus \(\mathrm{k}\) and modulus of rigidity \(\eta\) ? (A) \(\mathrm{Y}=[9 \eta \mathrm{k} /(\eta+3 \mathrm{k})]\) (B) \(\mathrm{Y}=[9 \mathrm{Yk} /(\mathrm{y}+3 \mathrm{k})]\) (C) \(\mathrm{Y}=[9 \eta \mathrm{k} /(3+\mathrm{k})]\) (D) \(\mathrm{Y}=[3 \eta \mathrm{k} /(9 \eta+\mathrm{k})]\)

The relation between surface tension T. Surface area \(\mathrm{A}\) and surface energy \(\mathrm{E}\) is given by. (A) \(\mathrm{T}=(\mathrm{E} / \mathrm{A})\) (B) \(\mathrm{T}=\mathrm{EA}\) (C) \(\mathrm{E}=(\mathrm{T} / \mathrm{A})\) (D) \(\mathrm{T}=(\mathrm{A} / \mathrm{E})\)

The value of poisson's ratio lies between......... (A) - 1 to \((1 / 2)\) (B) \(-(3 / 4)\) to \([(-1) / 2]\) (C) \(-(1 / 2)\) to 1 (D) 1 to 2
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