Heating Water with an Electric Heater: Temperature Rise Calculation
When we consider the use of an electric heater for heating water, we often need to determine the temperature rise required for different masses of water over varying time periods. This article will explore a specific scenario where we calculate the temperature rise of water when heated by an electric heater. We will use the principles of thermodynamics and the specific heat capacity of water to solve this problem.
Calculating Heat Gain and Temperature Rise for 0.5 kg of Water
Let's begin by examining the first scenario where 0.5 kg of water is heated from 15°C to 39°C in 3 minutes.
Mass of Water, m1: 0.5 kg Rise in Temperature, t1: 39°C - 15°C 24°C Specific Heat of Water, s: 4184 J/°C/kg Heat Gained by the Water, h1:The heat gained by the water can be calculated using the formula:
h1 m1 x s x t1
Plugging in the values:
h1 0.5 kg x 4184 J/°C/kg x 24°C 50208 J
This heat is produced by the heater in 3 minutes. Therefore, the heat produced per minute by the heater is:
Heat produced per minute 50208 J / 3 minutes 16736 J/minute
Calculating Heat Gain and Temperature Rise for 0.6 kg of Water over 8 Minutes
We now need to determine the temperature rise of 0.6 kg of water when heated for 8 minutes by the same heater.
Heat Produced by the Heater in 8 Minutes, h2: Mass of Water, m2: 0.6 kg Lets Assume t2°C be the rise in temperature for 0.6 kg of water:The heat gained by the water can be expressed as:
m2 x s x t2 h2
Using the heat produced per minute from the first scenario:
h2 16736 J/minute x 8 minutes 133888 J
Plugging in the values:
0.6 kg x 4184 J/°C/kg x t2 133888 J
Solving for t2:
t2 133888 J / (0.6 kg x 4184 J/°C/kg) 53.3°C
Simplified Method for Quick Calculation
An alternative and simpler way to calculate the temperature rise involves using a proportional relationship. We can use the following formula to estimate the temperature rise for the second scenario:
Temperature rise, t2 (0.6 kg / 0.5 kg) x (8 minutes / 3 minutes) x 24°C
Plugging in the values:
t2 (0.6 / 0.5) x (8 / 3) x 24°C 1.2 x 2.67 x 24°C 76.8°C
This quick method provides a close approximation for the temperature rise, simplifying the calculation without losing the essence of the problem.
Conclusion
The calculations demonstrate the practical application of specific heat capacity and the principles of thermodynamics to solve real-world problems related to heating water. Understanding these principles is crucial for various applications, from home heating systems to industrial processes.
By mastering these calculations, one can optimize the use of energy and resources in various heating applications.