PHYSICS HIGH SCHOOL

A student measured the specific heat of water to be 4.29 J/g.Co. Theliterature value of the specific heat of water is 4.18 J/g.Co. What was the
student's percent error? *
A) .0256%
B) 2.56%
C) 2.63%
D) .0263%

Answers

Answer 1

Answer:

Answer:

2.63 %.

Explanation:

Given that,

The calculated value of the specific heat of water is 4.29 J/g.C

Original value of specific heat of water is 4.18 J/g.C.

We need to find the student's percent error. The percentage error in any quantity is given by :

$P=\frac{|\text{original value-calculated value}|}{\text{original value}}* 100\n\nP=(4.29-4.18)/(4.18)* 100\n\nP=2.63\%$

So, the student's percent error is 2.63 %.

Answer 2

Answer:

The student's percent error in measuring the specificheat of water is 2.63%.

The specificheat is the amount of heat required to increase the temperature of 1 kg of mass by 1 degree Celsius.

Given:

Experimental Value = 4.29 J/g°C (student's measured value)

Accepted Value = 4.18 J/g°C (literature value)

The percent error can be calculated using the formula:

Percent Error = ((Experimental Value - Accepted Value) / Accepted Value)×100

Substitute the given values in the above formula and calculate the percenterror:

Percent Error = [(4.29 - 4.18) / 4.18] × 100

Percent Error = (0.11 / 4.18) × 100

Percent Error = 2.63%

The student's percent error in measuring the specificheat of water is approximately 2.63%.

Therefore, the student's percent error in measuring the specific heat of water is 2.63%.

To know more about the specificheat, click here:

brainly.com/question/31608647

#SPJ3

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Answers

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A student measured the specific heat of water to be 4.29 J/g.Co. Theliterature value of the specific heat of water is 4.18 J/g.Co. What was the
student's percent error? *
A) .0256%
B) 2.56%
C) 2.63%
D) .0263%