Water is pumped from a lower reservoir to a higher reservoir by a pump that prov

Step 1:
The mechanical power provided by the pump is 20 kW. We need to convert this power to the flow rate of water using the equation:
$P_{\text{mech}}=\rho gQH$
where:
- $P_{\text{mech}}$ is the mechanical power provided by the pump (20 kW)
- $\rho$ is the density of water (taken as $1000kg/m^3$)
- $g$ is the acceleration due to gravity (taken as $9.8m/s^2$)
- $Q$ is the flow rate of water ($0.03m^3/s$)
- $H$ is the height difference between the upper and lower reservoirs (45 m)
Solving for $P_{\text{mech}}$, we have:
$20\text{kW}=(1000{\text{kg/m}}^3)·(9.8{\text{m/s}}^2)·(0.03{\text{m}}^3/\text{s})·45\text{m}$
Step 2:
Now, we can calculate the value of $P_{\text{mech}}$:
$P_{\text{mech}}=(1000{\text{kg/m}}^3)·(9.8{\text{m/s}}^2)·(0.03{\text{m}}^3/\text{s})·45\text{m}=13230\text{W}$
Therefore, the mechanical power converted to thermal energy due to frictional effects during this process is 13,230 W or 13.23 kW.

Answer:
$-112300\text{W}$
Explanation:
The mechanical power provided by the pump is 20 kW, which we can express as $P_{\text{mech}}=20\text{kW}$.
The flow rate of water is given as $Q=0.03\left(\frac{{\text{m}}^3}{\text{s}}\right)$.
The height difference between the two reservoirs is 45 m, denoted as $h=45\text{m}$.
We can calculate the potential energy change of the water using the formula $PE=m·g·h$, where $m$ is the mass of water and $g$ is the acceleration due to gravity.
The mass of water can be calculated using the flow rate and the density of water, denoted as $\rho =1000\left(\frac{\text{kg}}{{\text{m}}^3}\right)$.
So, $m=\rho ·Q$.
The potential energy change can then be expressed as $PE=(\rho ·Q)·g·h$.
The mechanical power can be related to the potential energy change by the equation $P_{\text{mech}}=\frac{\text{d}PE}{\text{d}t}$.
Therefore, the power converted to thermal energy due to frictional effects, denoted as $P_{\text{thermal}}$, can be calculated as $P_{\text{thermal}}=P_{\text{mech}}-\frac{\text{d}PE}{\text{d}t}$.
Substituting the given values into the equations, we can calculate $P_{\text{thermal}}$.
Let's proceed with the calculations:
$\begin{array}{c}\hfill P_{\text{mech}}=20\text{kW}\\ \hfill Q=0.03\left(\frac{{\text{m}}^3}{\text{s}}\right)\\ \hfill h=45\text{m}\\ \hfill \rho =1000\left(\frac{\text{kg}}{{\text{m}}^3}\right)\end{array}$
First, we calculate the potential energy change:
$\begin{array}{c}\hfill PE=(\rho ·Q)·g·h=(1000\left(\frac{\text{kg}}{{\text{m}}^3}\right)·0.03\left(\frac{{\text{m}}^3}{\text{s}}\right))·9.8\left(\frac{\text{m}}{{\text{s}}^2}\right)·45\text{m}=132300\text{W}\end{array}$
Now, we can calculate the power converted to thermal energy:
$\begin{array}{c}\hfill P_{\text{thermal}}=P_{\text{mech}}-\frac{\text{d}PE}{\text{d}t}\\ \hfill =20\text{kW}-\frac{\text{d}(132300\text{W})}{\text{d}t}=20\text{kW}-132300\text{W/s}=-112300\text{W}\end{array}$
Therefore, the mechanical power converted to thermal energy due to frictional effects is $-112300\text{W}$.

Let's denote the flow rate of water as $Q=0.03{\text{m}}^3/\text{s}$ and the height difference between the reservoirs as $h=45\text{m}$. The mechanical power provided by the pump is $P_{\text{mech}}=20\text{kW}$.
The potential energy change of the water is given by the formula:
$\Delta PE=m·g·h$
where $m$ is the mass of water and $g$ is the acceleration due to gravity.
Since the flow rate is given, we can find the mass of water per second using the formula:
$m=Q·\rho$ where $\rho$ is the density of water.
Substituting the expression for $m$ into the potential energy formula, we have:
$\Delta PE=(Q·\rho )·g·h$
Now, let's find the power converted to thermal energy due to frictional effects. The total power provided by the pump is equal to the sum of the power used to increase the potential energy and the power converted to thermal energy:
$P_{\text{total}}=P_{\text{mech}}=P_{\text{PE}}+P_{\text{thermal}}$
The power used to increase the potential energy can be calculated as:
$P_{\text{PE}}=\Delta PE/t$ where $t$ is the time taken to pump the water from the lower reservoir to the higher reservoir.
Finally, we can find the power converted to thermal energy as:
$P_{\text{thermal}}=P_{\text{total}}-P_{\text{PE}}$
Now, let's calculate the values using the given data:
- Flow rate: $Q=0.03{\text{m}}^3/\text{s}$
- Height difference: $h=45\text{m}$
- Mechanical power: $P_{\text{mech}}=20\text{kW}$
The density of water, $\rho$, is approximately $1000{\text{kg/m}}^3$, and the acceleration due to gravity, $g$, is approximately $9.8{\text{m/s}}^2$.
First, let's calculate the potential energy change:
$\Delta PE=(Q·\rho )·g·h=(0.03{\text{m}}^3/\text{s}·1000{\text{kg/m}}^3)·(9.8{\text{m/s}}^2)·45\text{m}$
Substituting the values, we find:
$\Delta PE=13230\text{W}$
Next, let's calculate the power used to increase the potential energy:
$P_{\text{PE}}=\Delta PE/t$
Since the time $t$ is not given, we cannot calculate the exact value of $P_{\text{PE}}$ without it.
Finally, let's find the power converted to thermal energy:
$P_{\text{thermal}}=P_{\text{total}}-P_{\text{PE}}$
Substituting the given mechanical power $P_{\text{mech}}=20\text{kW}$ and the calculated $P_{\text{PE}}$, we can find $P_{\text{thermal}}$.