1 Calculate the lculate the amount of heat released when a 225 g sample of gold
ID: 508961 • Letter: 1
Question
1 Calculate the lculate the amount of heat released when a 225 g sample of gold cools from 860 C to32.5 C. The specific he of gold is 0.129J/g C. 2 Calculate the specific heat of an unknown metal fa 32.5 g sample loses 328Jof heat when it cools from 64.0 C to 16.0°C. 3 A 1.00 kg piece Cisplaced into 100.gof water at 10.0 C. The final temperature of metal at 52.0 capacity of 42.0 J/C, calculate the of the water is 31.4 C. Assuming a coffee cup calorimeter heat specific heat of the metal. The specific heat of water is 4.18 J/g"C.Explanation / Answer
1. given specific heat of gold = 0.129 J/g0C,
Mass of gold =225 g,
initial temperature 86 0C ,
Final emperature =32.5 0C
EQUATION : q(heat) = mass(gold) x specific heat capacity(gold) x change in temperature(solution)
= 225g x 0.129 J/g0C X (32.5 - 86)0C
=(-)1552.8 J where -ve sign indicates that heat is liberated.
2. Given mass of sample = 32.5 g
heat lost = (-) 328J
initial temperature 60 0C ,
Final emperature =16 0C
EQUATION : q(heat) = mass x specific heat capacity x change in temperature(solution)
SPECIFIC HEAT = heat (q)/ mass x change in temperature
= (-)328/32.5 x (16-60)
= 0.229 J/g 0C
3. GIVEN mass of metal piece = 1 kg =1000g
mass of water =100 g
initial temperature of water = 10 0C
final temperature of water =31.40C
initial temperature of metal = 520C
final temperature of METAL =31.40C
HEAT CAPACITY OF CALORIMETER =420 J/0C
specific heat of water =4.18 J/g0C
SPECIFIC HEAT OF METAL =?
q(metal) =-q (water)
specific heat of metal x mass of metal x temperature change of metal = (-)specific heat of water x mass of water x temperature change of water.
= specfic heat of metal x 1000g x (31.4-52) = (-)100 x 4.18 x (31.4-10)
specfic heat of metal X (-) 20600 g 0C = (-) 8945.2 J
SPECIFIC HEAT OF METAL = 8945.2J/ 20600 g 0C
= 0.434 J/ g 0C
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.