The temperature of a pot of water is 97.0 degeee C and the mass of the water is
ID: 1474654 • Letter: T
Question
The temperature of a pot of water is 97.0 degeee C and the mass of the water is 127 g. A block with a mass of 23.0 g and a temperature of 20.0 degree C is dropped into the water. The final temperature of the water and the block is 95.4 degree C. The specific heat of water is exactly 1 cal/g degree C. How much heat was lost by the water 36 8 cal 203 cal 1.60 cal 9580 cal How much heat was gained by the block 36.8 cal 75.4 cal 1730 cal 203 cal What is the specific heat of the block 0.117 cal/g degree C 352000 cal/g degree C 5.52 cal/g degree C 0.0212 cal/g degree CExplanation / Answer
a.how much heat gaind by water?
heat gained = mass of water x sp heat of water x change in temp of water
Q=MCT and if you shoud use c in cal/g C and mass in g so equation will be like:
Q= 127 *1cal/g* (95.4 - 97)=-203.2cal
b.if the system doesnt have heat leak the heat that water take is the heat that metal lost but it is negative so Qm=203.2cal
C. heat, q = m x sp heat x change in temp
heat lost by block = heat gained by water
m of block x sp heat of block x change in temp of block = mass of water x sp heat of water x change in temp of water
23 g x (sp heat) x (95.4 C- 20 C) = 127 g x (1 cal/gC) x (95.4 C- 97 C)
23 g x (sp heat) x ( 75.4C) = 127 cal/C x ( -1.6 C)
disregard the negative sign for the temp of water. the negative sign indicates that heat is lost. for mathematical equality, both sides of the equation should have same sign.
1734.2 g C X sp heat = 203.2cal
sp heat of metal = 0.117 cal/gC
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