Table of Contents
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The temperature of water in a beaker is 50°C, what is its value on the Fahrenheit scale? (122°F)
Normal human body temperature is 98.6°F, convert it into Celsius scale and Kelvin scale. (37°C,310K)
Calculate the increase in the length of an aluminum bar 2 m long when heated from 0°C to 20°C, if the thermal coefficient of linear expansion of aluminum is $2.5 \times 〖10〗^{-5} K^{-1}$. (0.1cm)
Solution: Original length of rod = $L_{(0=)}$ 2m
Initial temperature = $T_{(0=)}$ 0°C = 0 + 273 = 273 K
Final temperature = T = 20°C = 20 + 273 = 293 K
Change in temperature = $\triangle T$ = $T - T_{0}$ = 293 – 273 = 20 K
Coefficient of linear expansion of aluminum = α = $2.5 \times 10^{-5} K^{-1}$
Increase in volume $\triangle L$ = ?
$ \triangle L = α L_0 \triangle$ T
$\triangle L = 2.5 \times 10^{-5} \times20$
$\triangle L = 100 \times 10^{-5}$
$\triangle L = 0.001 m = 0.001\times$ 100 = 0.1cm
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A balloon contains $1.2m^{3}$ air at 15°C. Find its volume at 40°C. The thermal coefficient of volume expansion of air is $3.67\times〖10〗^{-3} K^{-1}$. $(1.3m^{(3)}$
Solution: Original volume = $V_{(0=)} 1.2m^{3}$
Initial temperature = $T_{0}$ = 15°C = 15 + 273 = 288 K
Final temperature = T = 40°C = 40 + 273 = 313 K
Change in temperature = $\triangle T$ = T - $T_{0}$ = 313 – 288 = 25 K
Coefficient of volume expansion of air $\beta$ = $3.67 \times 10^{-3} K^{-1}$
Volume = V = ?
V = $V_{0} (1+ \beta \triangle T)$
V = $1.2 (1 + 3.67 \times 10^{-3}\times25)$= $1.2(1+ 91.75 \times 10^{-3})$
= 1.2(1 + 0.09175) = $1.2\times 1.09175$
V = $1.3m^{3}$
How much heat is required to increase the temperature of 0.5 kg of water from 10°C to 65°C? (115500 J)
Solution: Mass of water = m = 0.5 kg
Initial temperature = $T_{1}$ = 10°C = 10 + 273 = 283 K
Final temperature = $T_{2}$ = 65°C = 65 + 273 = 338 K
Change in temperature = $\triangle$ T = $T_{2} - T_{1}$ = 338 – 283 = 55 K
Heat =$ \triangle Q$ = ?
$\triangle Q$ = $mc \triangle T$
$\triangle Q = 0.5\times 2400 \times55$
$\triangle Q$ = 115500J
An electric heater supplies heat at the rate of 1000J per second. How much time is required to raise the temperature of 200 g of water from 20°C to 90°C? (58.8 s)
Solution: Power = P = 1000 $Js^{-1}$
Mass of water = m = 200 g =$\frac{200}{1000}$ = 0.2 kg
Initial temperature = $T_{2}$= 20°C = 20 + 273 = 293 K
Final temperature = $T_{1}$= 90°C = 90 + 273 = 363 K
Change in temperature = $\triangle T$ = $T_{2} - T_{1}$ = 363 – 293 = 70 K
Specific heat of water = c = $4200 J〖kg〗^{-1} K^{-1}$
Time = t = ?
P =$\frac{w}{t}$
Or P =$\frac{Q}{t}$
Or P $\times t$ = Q
Or P $\times t$ = $mc \triangle T$
Or t=$\frac{(mc \triangle T)}{P}$
t = $\frac{(0.2\times 4200 \times 70)}{1000}$= 58.8 s
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Solution: Amount of heat required to melt ice = 50000J
Latent heat of fusion of ice = = $336000 J 〖kg〗^{-1}$
Amount of ice = m =?
$ \triangle Q_{f}$= $m H_{f}$
Or m = $\frac{(\triangle Q_{f} )}{(H_{f})}$
m =$\frac{50000}{336000}$ = 0.1488 kg
= $ 0.1488 \times 1000$ =$\frac{1488}{1000} \times $1000 = $148.8 g \approx 149 g$
Find the quantity of heat needed to melt 100g of ice at -10°C into the water at 10°C.
(39900 J)
(Note: Specific heat of ice is $2100 J〖kg〗^{-1} K^{-1}$, the specific heat of water is $4200 J〖kg〗^{-1} K^{-1}$, Latent heat of fusion of ice is 336000 J$〖kg〗^{-1})$.
How much heat is required to change 100g of water at 100°C into steam? (Latent heat of vaporization of water is $2.26 \times 〖10〗^{6} J〖kg〗^{-1}$. $(2.26\times 〖10〗^{5}J)$
Solution: Mass of water = m = 100 g = $\frac{100}{1000}$ = 0.1 kg
Latent heat of vaporization of water = $H_{v}$ = $2.26 \times 10^{6} Jkg^{-1}$
Heat required = $\triangle Q_{v}$= ?
$\triangle Q_{v}= mH_{v}$
$\triangle Q_{v}$= $0.1\times 2.26 \times 10^{6}$ = $0.226 \times 10^{6}$ = $\frac{226}{1000} \times 10^{6}$
= $2.26 \times 10^{-1} \times 10^{6}$ = $2.26 \times 10^{5}$J
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