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# In the laboratory a student finds that it takes 82-2 Joules to increas.docx

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# In the laboratory a student finds that it takes 82-2 Joules to increas.docx

In the laboratory a student finds that it takes 82.2 Joules to increase the temperature of 11.8 grams of solid iron from 21.2 to 37.9 degrees Celsius. The specific heat of iron calculated from her data is ____J/gÂ°C.
Solution
The correct answer is 0.417 J/g o C
Heat gain equation can be written as Q Heat Gained = (m iron ) x (C iron ) x (delta T)
where Q Heat Gained is the energy gained by iron = 82.2 J
m iron is the mass of iron = 11.8 g
C iron = specific heat capacity of iron
delta T = change in temperature = T final - T initial = 37.9 o C - 21.2 o C = 16.7 o C
C iron = (Q Heat Gained ) / [(m iron ) x (delta T)]
C iron = (82.2 J) / [(11.8 g) x (16.7 o C)]
or C iron = 0.417 J/g o C
.

In the laboratory a student finds that it takes 82.2 Joules to increase the temperature of 11.8 grams of solid iron from 21.2 to 37.9 degrees Celsius. The specific heat of iron calculated from her data is ____J/gÂ°C.
Solution
The correct answer is 0.417 J/g o C
Heat gain equation can be written as Q Heat Gained = (m iron ) x (C iron ) x (delta T)
where Q Heat Gained is the energy gained by iron = 82.2 J
m iron is the mass of iron = 11.8 g
C iron = specific heat capacity of iron
delta T = change in temperature = T final - T initial = 37.9 o C - 21.2 o C = 16.7 o C
C iron = (Q Heat Gained ) / [(m iron ) x (delta T)]
C iron = (82.2 J) / [(11.8 g) x (16.7 o C)]
or C iron = 0.417 J/g o C
.

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### In the laboratory a student finds that it takes 82-2 Joules to increas.docx

1. 1. In the laboratory a student finds that it takes 82.2 Joules to increase the temperature of 11.8 grams of solid iron from 21.2 to 37.9 degrees Celsius. The specific heat of iron calculated from her data is ____J/gÂ°C. Solution The correct answer is 0.417 J/g o C Heat gain equation can be written as Q Heat Gained = (m iron ) x (C iron ) x (delta T) where Q Heat Gained is the energy gained by iron = 82.2 J m iron is the mass of iron = 11.8 g C iron = specific heat capacity of iron delta T = change in temperature = T final - T initial = 37.9 o C - 21.2 o C = 16.7 o C C iron = (Q Heat Gained ) / [(m iron ) x (delta T)] C iron = (82.2 J) / [(11.8 g) x (16.7 o C)] or C iron = 0.417 J/g o C