Use below values as needed in the following calculations: Substance Specific Hea
ID: 1043781 • Letter: U
Question
Use below values as needed in the following calculations: Substance Specific Heat (J/goc) H20 () 4.184 Aluminum (s) 0.89 Iron (s) 0.45 4. How much heat is required to warm 275 g of water from 76 oC to 87 oC? 5. PC13 is a compound used to manufacture pesticides. A reaction requires that 96.7 g of PC13 be raised from 31.7oC to 69.20C. How much energy will this require given that the specific heat of PC13 is 0.874 J/g oC? 6. A quantity of water is heated from 25.0 oC to 36.4 oC by ab sorbing 325 J of heat energy. W hat is the mass of the water? 7.A 500. g sample of an unknown metal releases 6.4 x 102 Jas it cools from 55.0 oC to 25.0 oC. What is the specific heat of the sample? 8. How much heat is lost when a 15.34 g piece of Iron temperature drops from 90.0 °C to 81.0 °C? 9. A 250 g sample of water with an initial temperature of 98.8 oC loses 7500 joules of heat. What is the final temperature of the water? (Remember, final temp initial temp change in temp)Explanation / Answer
4) Heat required for raising temperature = Q = m X s X delta(t)
m = mass of water = 275 g
s = specific heat = 4.184
delta (t) = change in temperature = 87- 76 = 11 0C
Q = 275 X 4.184 X 11 = 12610.58 J = 12.611 kJ
5) Q = m X s X delta(t)
m = mass =96.7g
s = specific heat = 0.874
delta (t) = 69.2 - 31.7 = 37.5
Q = 96.7 X 0.874 X 37.5 = 3169.34 Joules
6) Q = m X s X delta(t)
m = ?
Q = 325 J
delta(t) = 36.4 - 25 = 11.4 0C
s = 4.184
Putting values
325 = m X 4.184 X 11.4
m = mass = 6.184 grams
7 ) Q = m X s X delta(t)
m = 500g
Q = 640 J
delta(t) = 55-25 = 30 0C
s = ?
Putting values
640 = 500 X s X 30
s = 0.0427 J /0C g
8) Q = m X s X delta(t)
m =15.34 g
Q = ?
delta(t) = 90-81 = 9 0C
s = 0.45
Q = 15.34 X 0.45 X 9 = 62.127 g
9) Q = m X s X delta(t)
m = 250g
Q = 7500 J
delta(t) = ?
s = 4.184
Putting values
7500 = 250 X 4.184 X delta (t)
delta (t) = change in temperature = 7.17 0C
Final temperature = Initial temperature - change in temperature = 98.8 0C - 7.17 0C = 91.63 0C
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