How long does a `59 kw` water heater take to raise the temperature of `150L` of water from `21^(@)C` to `38^(@)C` (in min)

Explanation

To solve the problem of how long a 59 kW water heater takes to raise the temperature of 150 liters of water from 21°C to 38°C, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Power of the heater, \( P = 59 \, \text{kW} = 59 \times 10^3 \, \text{W} \)
- Volume of water, \( V = 150 \, \text{L} \)
- Temperature change, \( \Delta T = 38°C - 21°C = 17°C \)

2. **Convert Volume to Mass:**
- Since the density of water is approximately \( 1 \, \text{kg/L} \), the mass of water can be calculated as:
\[ m = V = 150 \, \text{kg} \]

3. **Calculate the Heat Required to Raise the Temperature:**
- The formula for heat absorbed by the water is given by:
\[ Q = m \cdot c \cdot \Delta T \]
- Where \( c \) (specific heat capacity of water) is approximately \( 4200 \, \text{J/kg°C} \).
- Substituting the values:
\[ Q = 150 \, \text{kg} \cdot 4200 \, \text{J/kg°C} \cdot 17°C \]
- Calculate \( Q \):
\[ Q = 150 \cdot 4200 \cdot 17 = 107,100,000 \, \text{J} \]

4. **Relate Heat to Power and Time:**
- The relationship between heat, power, and time is given by:
\[ Q = P \cdot t \]
- Rearranging for time \( t \):
\[ t = \frac{Q}{P} \]
- Substitute \( Q \) and \( P \):
\[ t = \frac{107,100,000 \, \text{J}}{59 \times 10^3 \, \text{W}} \]

5. **Calculate Time in Seconds:**
- Performing the calculation:
\[ t = \frac{107,100,000}{59,000} \approx 1815.25 \, \text{s} \]

6. **Convert Time to Minutes:**
- Since the question asks for time in minutes, convert seconds to minutes:
\[ t_{\text{min}} = \frac{1815.25}{60} \approx 30.25 \, \text{min} \]

### Final Answer:
The time taken by the 59 kW water heater to raise the temperature of 150 liters of water from 21°C to 38°C is approximately **30.25 minutes**.

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