Chap. 13, #2. (I) How does the number of atoms in a 26.5-gram gold ring compare to the number in a silver ring of the same mass?
The number of atoms in a mole of any substance is the same, equal to the Avogadro number. Since the Atomic mass of Gold (79) is higher than the atomic mass of Silver (47) the number of moles of gold is less than the number of moles of Silver in the same quantity of substance. It implies the number of atoms of gold is smaller than the number of atoms of silver in the same mass of substance.
Chap. 13, #4. (I) Among the highest and lowest temperatures recorded are 136°F in the Libyan desert and in Antarctica. What are these temperatures on the Celsius scale?
$C = (F-32)*5/9$
$136 F = (136-32)*5/9 =57.78 C$
$-129 F = (-129-32)*5/9 =-89.44 C$
Chap. 13, #12. (II) A quartz sphere is 8.75 cm in diameter. What will be its change in volume if it is heated from 30°C to 200°C?
Coefficient of linear dilatation of quartz is $\alpha =0.33*10^-6 Celsius^-1$
The volumetric coefficient is $\gamma = 3*alpha = 1*10^-6 Celsius^-1$
$(V-V0)/V0 = \gamma(t- t0)$
$V-V0 = V0*\gamma*(t-t0)$
$V0 = 4*\pi*R^3/3 = 4*\pi*4.375^3/3 =350.77 cm^3$
$V-V0 = 350.77*1*10^-6*(200-30) =0.0596 cm^3$
Chap. 13, #42. (I) How many moles of water are there in 1.000 L? How many molecules?
Molecular mass of water is $M(H_2O) =2*1+16 =18 grams$
1 liter of water wight 1000 g (density of water is 1000 g/L)
No of moles is $N = 1000/18 = 55.55 moles$
Chap. 14, #6. (II) A small immersion heater is rated at 350 W. Estimate how long it will take to heat a cup of soup (assume this is 250 mL of water) from 20°C to 60°C.
The total heat is $Q = Power*time = P*t$
On the other hand $Q = m*c*(t-t_0) = 250*4.18*(60-20) =41800 J$
$c(water)= 4.18 J/gram$
$time = Q/P = 41800/350 =119.42 seconds$
Chap. 14, #12. (II) What will be the equilibrium temperature when a 245-g block of copper at 285°C is placed in a 145-g aluminum calorimeter cup containing 825 g of water at 12.0°C?
The heat released by copper is equal to the heat taken by water and aluminum container.
$m(copper)*c(copper)*(285-t) = (m(water)*c(water) + m(Al)*c(Al))*(t-12)$
$c(Al) = 0.897 J/(g*K)$
$c(Cu) =0.385 J/(g*K)$
$c(water) =4.18 J/g*K$
$245*0.385*(285-t) = (825*4.18 +145*0.897)(t-12)$
$94.325*285 -94.325*t = 3578.565 *t -3578.565*12$
$3672.89*t = 69825.4$
$t = 19.01 Celsius$
Chap. 14, #14. (II) A 215-g sample of a substance is heated to 330°C and then plunged into a 105-g aluminum calorimeter cup containing 165 g of water and a 17-g glass thermometer at 12.5°C. The final temperature is 35.0°C. What is the specific heat of the substance? (Assume no water boils away.)
$m(sample)*c_x*(330-35) = m(Al)*c(Al)*(35-12.5) + $
$+m(water)*c(water)*(35-12.5) + m(glass)*c(glass)*(35-12,5)$
$c(water) =4.18 J/(g*K)$
$c(glass) = 0.84 J/(g*K)$
$215*c_x*295 =105*0.897*22.5 +165*4.18*22,5 +17*0.84*22.5$
$63425*c_x = 17958.7$
$c_x = 0.283 J/(g*K)$
Chap. 14, #41. (II) A 100-W light bulb generates 95 W of heat, which is dissipated through a glass bulb that has a radius of 3.0 cm and is 1.0 mm thick. What is the difference in temperature between the inner and outer surfaces of the glass?
Transferred quantity of heat is $Q = k*A*∆T/x$
Temperature difference between inner and outer surface is $∆T = Q*x/(k*A)$
Heat is $Q =Power*time =95 J/second$
x is the thickness of the glass $x = 0.001 m$
Surface area is A = $4*\pi*0.03^2 m^2$
$k = 1.1 W/(m-ºK)$ for glass
Therefore $∆T = 7.64 ºK$