In this scenario, we are examining the process of heating water using a hot plate rated at 200 watts. The goal is to determine the time required for the water to increase in temperature from 10 degrees Celsius to 90 degrees Celsius. To solve this, we will utilize the relationship between energy, power, and time.

First, we recognize that the energy transferred to the water, denoted as \( Q \), can be calculated using the specific heat formula:

\[ Q = m \cdot c \cdot \Delta T \]

Here, \( m \) represents the mass of the water, \( c \) is the specific heat capacity of water (approximately 4186 J/kg·°C), and \( \Delta T \) is the change in temperature. In this case, the change in temperature is:

\[ \Delta T = 90°C - 10°C = 80°C \]

Next, we need to find the time \( \Delta t \) it takes for this energy transfer to occur. The power equation relates power \( P \) to work done (or energy transferred) over time:

\[ P = \frac{W}{\Delta t} \]

Since work (or energy) can be represented as \( Q \), we can rearrange this equation to solve for time:

\[ \Delta t = \frac{Q}{P} \]

Substituting the expression for \( Q \) into this equation gives us:

\[ \Delta t = \frac{m \cdot c \cdot \Delta T}{P} \]

Now, we can plug in the values:

Mass of water, \( m = 0.3 \) kg
Specific heat of water, \( c = 4186 \) J/kg·°C
Change in temperature, \( \Delta T = 80 \) °C
Power of the hot plate, \( P = 200 \) W

Calculating \( Q \):

\[ Q = 0.3 \, \text{kg} \cdot 4186 \, \text{J/kg·°C} \cdot 80 \, \text{°C} = 100,464 \, \text{J} \]

Now substituting \( Q \) into the time equation:

\[ \Delta t = \frac{100,464 \, \text{J}}{200 \, \text{W}} = 502.32 \, \text{s} \]

This results in approximately 500 seconds, which is equivalent to about 8.3 minutes. Therefore, it will take roughly 8 and a half minutes to heat the water from 10 degrees Celsius to 90 degrees Celsius.

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