A piece of iron (400. g) is heated to a temperature of 330.°C and is then dropped into a beaker containing 1.00 kg of water. If the original temperature of the water was 2.0°C and that of the mixture was 32.8°C, what is the calculated specific heat capacity of iron?
P.S. What I did was use the formula q=ms∆t. I manipulated it so that I could have it be s=q/(m∆t). My problem comes in when substituting the letters for numbers. I don't have any idea where to get the quantitative information for q.
Thermochemistry question.?
2016-07-17 8:58 pm
回答 (2)
2016-07-17 9:08 pm
✔ 最佳答案
specific heat capacity of water = 4.186 J/(g °C)q = m s ΔT
Heat lost by the iron = 400 × s × (330 - 32.8) J
Heat gained by the water = 1000 × 4.186 × (32.8 - 2.0) J
Heat lost by the iron = Heat gained by the water
400 × s × (330 - 32.8) = 1000 × 4.186 × (32.8 - 2.0)
400 × s × (330 - 32.8) = 1000 × 4.186 × (32.8 - 2.0)
Specific heat capacity of iron, s = 1.08 J/(g °C)
2016-07-17 9:13 pm
In order to answer this question you need to know the specific heat of water.
The source below says 4.184 J/g·°C.
(4.184 J/g·°C) x (1000 g) x (32.8 - 2.0) °C = 128867.2 J gained by the water
(128867.2 J) / (400 g) / (330 - 32.8) °C = 1.08 J/g·°C
[But that's a long way from the accepted value for the specific heat of iron.]
The source below says 4.184 J/g·°C.
(4.184 J/g·°C) x (1000 g) x (32.8 - 2.0) °C = 128867.2 J gained by the water
(128867.2 J) / (400 g) / (330 - 32.8) °C = 1.08 J/g·°C
[But that's a long way from the accepted value for the specific heat of iron.]
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