How Long Does It Take to Heat Water in a Kettle with a Water Equivalent of 150g?

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Homework Help Overview

The discussion revolves around the heating of water in a kettle, specifically calculating the time required to raise the temperature of water from 15 degrees Celsius to 90 degrees Celsius. The kettle's water equivalent is noted to be 150g, and the problem involves understanding the power input, resistance, and efficiency of the kettle's heating element.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss calculations involving power, resistance, and efficiency, with attempts to reconcile their results with textbook answers. Questions arise regarding the assumptions made about the mass of water and the kettle's water equivalent, particularly in the context of heating a second litre of water immediately after the first.

Discussion Status

The discussion is ongoing, with participants providing clues and prompting each other to reconsider their assumptions. There is an exploration of the relationship between the heating times for one litre versus two litres of water, and how the kettle's capacity and efficiency affect these calculations.

Contextual Notes

Participants are working under the assumption that 1 litre of water equals 1 kg and are considering the implications of the kettle's water equivalent in their calculations. There is a noted discrepancy between personal calculations and textbook values, which is prompting further inquiry.

ande
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1. The element of a heater kettle has a constant resistance of 105 ohms and applied voltage of 240V.

1. Calculate the time taken to raise the temperature of water from 15 degrees Celsius to 90 assuming that. 80% of the power input to the kettle is usefully employed. I worked this out and got 11 min 56sec which is the same answer that appears on the textbook

2. If the water equivalent of the kettle is 150g, find how long it will take to raise a second litre of. Water through the same temp range immediately after the first time. I get 10min 08sec but the textbook says 10min 58sec. Please shed some light
2. Q= mcdt P= V^2/R and efficiency = Po/Pi3. I attempted the question by subtracting the equivalent mass from the mass of water thus giving a net mass of 0.85 but the answer I got doesn't correspond with the textbook. Assume 1L = 1kg
 
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ande said:
a second litre
This is a clue.
ande said:
giving a net mass of 0.85
Think about the assumption you made to get this number, then try it again.
 
Bystander said:
This is a clue.

Think about the assumption you made to get this number, then try it again.
Correct me of I'm wrong but if the second litre is heated immediately after the first, 150g of the system is theoretically at 90 degrees correct? You'll have to pardon my reasoning, I'm a first year electrical engineering student.
 
Figure the capacity of the kettle is at least two liters. If you can heat one liter plus the kettle in time x, at whatever efficiency, if the efficiency is constant, you can heat two liters plus the kettle in what time? What is the difference between the two times?
 

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