Equal 0.390-kg masses of lead and tin at 60.0°C are placed in 0.80 kg of water a
ID: 1511322 • Letter: E
Question
Equal 0.390-kg masses of lead and tin at 60.0°C are placed in 0.80 kg of water at 17.3°C.
(a) What is the equilibrium temperature of the system?
°C
(b) If an alloy is half lead and half tin by mass, what specific heat would you anticipate for the alloy?
J/kg · °C
(c) How many atoms of tin NSn are in 0.390 kg of tin, and how many atoms of leadNPb are in 0.390 kg of lead?
(d) Divide the number NSn of tin atoms by the number NPb of lead atoms and compare this ratio with the specific heat of tin divided by the specific heat of lead. What conclusion can be drawn?
Explanation / Answer
a) Here, mt * ct * (T - Tt) + ml * cl * (T - Tl) + mw * cw * (T - Tw) = 0
=> 0.390 * 210 * (T-60) + 0.390 * 128 * (T-60) + 0.80 * 4186 * (T - 17.3) = 0
=> 3480.62T = 65843.44
=> T = 18.917 oC --------------------> equilibrium temperature of the system
b) specific heat would anticipate for the alloy = 167 J/kg · °C
c) atoms of tin NSn are in 0.390 kg of tin = (390/118.71) * 6.023 * 1023
= 1.978 * 1024 atoms
atoms of leadNPb are in 0.390 kg of lead = (390/207.2) * 6.023 * 1023
= 1.1336 * 1024 atoms
d) Ratio of number NSn of tin atoms by the number NPb of lead atoms = (1.978 * 1024)/(1.1336 * 1024)
= 1.7448
specific heat of tin divided by the specific heat of lead = 210/128
= 1.64
=> Both ratios are approximately same .
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