Equal 0.390-kg masses of lead and tin at 60.0°C are placed in 0.80 kg of water a

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 lead NPb 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

Sp hat of water = 4186 J/kg K

Sp heat of Tin = 2180 J/kg k

Sp heat of lead = 1280 J/kg k

part A "

at equillirium

(0.39 * 2180 * (60-T)) + (0.39 * 128 * (60-T)) = 1 * 4180 * (T-20)

T = 27.1 deg C

----------------------------

part B :

for an alloy half tin and alf elad

sg heat = Sa = (SL + ST)/2

Sa = 173 J/kg K

-----------

part C:

NT = 20.3 *10^6

NL = 11.63*10^26

AVagardro No. = 6.023*10^23

Molar mass of tin = 118.7 g/mol

NT/NL = 1.75

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