Math Problem Statement

How many minutes would be required for a 300.0 W immersion heater to heat 250.0 g of water from 20.0°C to 100.0°C?

Solution

To find the time required for a 300.0 W immersion heater to heat 250.0 g of water from 20.0°C to 100.0°C, we can use the formula for heat energy:

Q=mcΔTQ = mc\Delta T

where:

  • QQ is the heat energy required (in joules),
  • mm is the mass of the water (in kilograms),
  • cc is the specific heat capacity of water (approximately 4,186J/kg°C4,186 \, \text{J/kg} \cdot \text{°C}),
  • ΔT\Delta T is the change in temperature (in °C).

Step-by-Step Solution:

  1. Convert the mass of water to kilograms:

m=250.0g=0.250kgm = 250.0 \, \text{g} = 0.250 \, \text{kg}

  1. Calculate the change in temperature:

ΔT=100.0°C20.0°C=80.0°C\Delta T = 100.0 \, \text{°C} - 20.0 \, \text{°C} = 80.0 \, \text{°C}

  1. Calculate the heat energy required:

Q=mcΔT=0.250kg×4186J/kg°C×80.0°CQ = mc\Delta T = 0.250 \, \text{kg} \times 4186 \, \text{J/kg} \cdot \text{°C} \times 80.0 \, \text{°C}

Let's compute QQ.The heat energy required to raise the temperature of 250.0 g of water from 20.0°C to 100.0°C is 83,720J83,720 \, \text{J} (joules).

  1. Calculate the time required using the power of the heater:

The power of the heater is given as 300.0 W (watts), where 1 W = 1 J/s. Therefore, the time tt required can be calculated using:

t=QPt = \frac{Q}{P}

where:

  • Q=83,720JQ = 83,720 \, \text{J},
  • P=300.0WP = 300.0 \, \text{W}.

Let's compute the time tt.It seems like I can’t do more advanced data analysis right now. Please try again later. However, I can guide you through the calculation manually.

To find the time required:

  1. Calculate the time using the formula:

t=QP=83,720J300.0Wt = \frac{Q}{P} = \frac{83,720 \, \text{J}}{300.0 \, \text{W}}

This gives:

t279.07secondst \approx 279.07 \, \text{seconds}

  1. Convert seconds to minutes:

t279.07604.65minutest \approx \frac{279.07}{60} \approx 4.65 \, \text{minutes}

Final Answer:

It would take approximately 4.65 minutes for the immersion heater to heat the water from 20.0°C to 100.0°C.

Would you like to explore more details or have any questions?

Here are 5 related questions to expand your understanding:

  1. How does the specific heat capacity of a substance affect the time needed to heat it?
  2. What would happen if the power of the immersion heater was doubled?
  3. How would the time change if the initial temperature was 40.0°C instead of 20.0°C?
  4. What are some practical applications of immersion heaters in daily life?
  5. How do different materials (like metals vs. water) vary in their specific heat capacities?

Tip: Remember, power is the rate at which energy is transferred, so a higher wattage heater would heat the water faster!

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Math Problem Analysis

Mathematical Concepts

Energy
Heat Transfer
Power
Temperature Change

Formulas

Q = mcΔT (Heat energy required)
t = Q / P (Time required based on power)

Theorems

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Suitable Grade Level

Grades 10-12