Why Is My Calculation of Heat Energy Incorrect?

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The discussion centers on calculating the heat energy needed to raise the temperature of a glass beaker and water. The original poster attempted to calculate the heat energy but received an incorrect answer, expecting 6.29 x 10^4. Participants noted potential errors in unit conversion between grams and kilograms, which could account for discrepancies in the calculations. Additionally, there was a misreading of a handwritten exponent that contributed to confusion. Clarifying these points may help resolve the calculation issue.
Noawun
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Homework Statement
A glass beaker of mass 215g contains 145g of water at 18.5 degrees Celsius. If the specific heat capacity of glass is 8.4 x 10^2 J kg^-1 K^-1, how much heat energy would need to be supplied to raise the temperature of the glass and water to 98.5 degrees Celsius?
Relevant Equations
Q=mc\Delta T
At first, I tried to calculate the heat energy required by doing this:
IMG_DB51C510AD86-1.jpeg

I realized I should calculate heat energy separately instead of grouping glass and water together so I did this:

IMG_729D7A630229-1.jpeg


But the answer is supposed to be 6.29 x 10^4.

I don't know how to solve this. Can anyone help please? Thank you
 
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Noawun said:
Homework Statement:: A glass beaker of mass 215g contains 145g of water at 18.5 degrees Celsius. If the specific heat capacity of glass is 8.4 x 10^2 J kg^-1 K^-1, how much heat energy would need to be supplied to raise the temperature of the glass and water to 98.5 degrees Celsius?
Relevant Equations:: Q=mc\Delta T

At first, I tried to calculate the heat energy required by doing this:
View attachment 295998
I realized I should calculate heat energy separately instead of grouping glass and water together so I did this:

View attachment 295999

But the answer is supposed to be 6.29 x 10^4.

I don't know how to solve this. Can anyone help please? Thank you
I get 63168, rounding to 63200. I suspect someone made the error of rounding intermediate values.
 
It also looks like he's mixed units (g vers kg) which gives the 10^3 difference in his answer? (I didn't check his work in detail, though)
 
berkeman said:
It also looks like he's mixed units (g vers kg) which gives the 10^3 difference in his answer? (I didn't check his work in detail, though)
Well spotted! Looking only at the last line, I misread the handwritten exponent 7, with its continental centre line and the graph paper line at the left, as a four.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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