A small electric immersion heater is used to heat 170g of water for a cup of instant coffee. The heater is labeled "180 watts" ( it converts electrical enery to thermal energy at this rate ). Calculate the time required to bring all this water from `23.0^(@)C` to `100^(@)C`, ignoring any heat losses.

Answer

To solve the problem of calculating the time required to heat 170g of water from 23.0°C to 100.0°C using a 180-watt immersion heater, we can follow these steps:

### Step 1: Identify the given values
- Mass of water (m) = 170 g = 170 × 10^(-3) kg = 0.170 kg
- Initial temperature (T_initial) = 23.0°C
- Final temperature (T_final) = 100.0°C
- Power of the heater (P) = 180 watts
- Specific heat capacity of water (c) = 4.2 × 10^3 J/(kg·°C)

### Step 2: Calculate the temperature change (ΔT)
\[ \Delta T = T_{\text{final}} - T_{\text{initial}} = 100.0°C - 23.0°C = 77.0°C \]

### Step 3: Calculate the heat required (Q)
Using the formula for heat transfer:
\[ Q = m \cdot c \cdot \Delta T \]
Substituting the values:
\[ Q = 0.170 \, \text{kg} \cdot 4.2 \times 10^3 \, \text{J/(kg·°C)} \cdot 77.0 \, \text{°C} \]
\[ Q = 0.170 \cdot 4.2 \times 10^3 \cdot 77.0 \]
\[ Q = 0.170 \cdot 4.2 \cdot 77.0 \times 10^3 \]
\[ Q = 0.170 \cdot 323.4 \times 10^3 \]
\[ Q = 54.978 \times 10^3 \, \text{J} \approx 54978 \, \text{J} \]

### Step 4: Relate heat to power and time
The power of the heater is defined as the rate of heat transfer:
\[ P = \frac{Q}{t} \]
Rearranging to find time (t):
\[ t = \frac{Q}{P} \]
Substituting the values:
\[ t = \frac{54978 \, \text{J}}{180 \, \text{W}} \]
\[ t \approx 305.43 \, \text{s} \]

### Step 5: Convert time to minutes
To convert seconds to minutes:
\[ t_{\text{minutes}} = \frac{t}{60} = \frac{305.43}{60} \approx 5.09 \, \text{minutes} \]

### Final Answer
The time required to heat the water from 23.0°C to 100.0°C is approximately **305.43 seconds** or **5.09 minutes**.

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