The temperature of water in a beaker is 50°C, what is its value on the Fahrenheit scale? (122°F)

Difficulty: Easy
Solution:   Temperature in Celsius scale = C = 50°C
The temperature on the Fahrenheit scale = F =?
F = 1.8C + 32
F = $1.8 \times$ 50 + 32 = 90 + 32
F = 122°F

Normal human body temperature is 98.6°F, convert it into Celsius scale and Kelvin scale.                                                                                                            (37°C,310K)

Difficulty: Easy
Solution:
The temperature on Fahrenheit scale = 98.6°F
The temperature on the Celsius scale =?
The temperature on the Kelvin scale =?

F = 1.8C + 32
1.8C = F – 32
1.8C = 98.6 -32
1.8C = 66.6
C =$\frac{(66.6)}{(1.8)}$ = 37°C

T(K) = C + 273
T(K) = 37 + 273
T(K) = 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)

Difficulty: Easy

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

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 $3.67\times〖10〗^{-3} K^{-1}$. $(1.3m^{(3)}$

Difficulty: Easy

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)

Difficulty: Easy

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)

Difficulty: Medium

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

How much ice will melt by 5000 J of heat? Latent heat of fusion of ice = $336000 J 〖kg〗^{-1}$.                                                                                                                       (149 g )

Difficulty: Easy

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$〖kg〗^{-1})$.

Difficulty: Medium
Solution:   Mass of ice = m = 100 g = $\frac{100}{1000}$  = 0.1 kg
Specific heat of ice = c1 = $2100 J〖kg〗^{-1} K^{-1}$
Latent heat of fusion of ice = L = $336000 J〖kg〗^{-1}$
Specific heat of water = c = $4200 J〖kg〗^{-1} K^{-1}$
Quantity of heat required = Q =?

Case I:
Heat gained by ice from -10°C to 0°C
Q1 = $mc \triangle T$
Q1 = $0.1\times 2100 \times10$ = 2100 J

Case II:
Heat required for ice to melt = Q2 = mL
= $0.1\times 336000$
Q2  = 33600 J

Case III:
The heat is required to raise the temperature of water from 0°C to 10°C
Q3= $mc \triangle T$
Q3 = $0.1\times 4200 \times 10$ = 4200 J
Total heat required = Q = Q1 + Q2 + Q3
Q = 2100 + 33600 + 4200
Q = 39900 J

How much heat is required to change 100g of water at 100°C into steam? (Latent heat of vaporization of water is $2.26 \times 〖10〗^{6} J〖kg〗^{-1}$.                $(2.26\times 〖10〗^{5}J)$

Difficulty: Easy

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

Find the temperature of the water after passing 5 g of steam at 100°C through 500 g of water at 10°C.         (16.2°C)
(Note: Specific heat of water is $4200 J〖kg〗^{-1} K^{-1}$, Latent heat of vaporization of water is $2.26 \times 〖10〗^{6} Jkg^{-1})$.
Difficulty: Hard
Solution:    Mass of stream = $m_{1}= 5 g$ =$\frac{5}{1000}$ kg = 0.005 kg
Temperature of stream = $T_{1}$= 100°C
Mass of water = $m_{2}$ = 0.5 kg
Temperature of water = $T_{2}$= 10°C
Final temperature = $T_{3}$= ?

Case I:
Latent heat lost by stream = Q1 = mL
Q1 = $0.005\times 2.26 \times 10^{6}$ = $11.3 \times 10^{3}$ = 11300 J

Case II:
Heat lost by stream to attain final temperature Q2 = $m_{1} c \triangle T$
Q2 = $0.005\times 4200 \times (100- T_{3})$
Q2 = $21(100- T_{3})$

Case III:
Heat gained by water Q3 = $m_{2} c \triangle T$
Q3 = $0.5\times 4200\times (T_{3} - 10)$
Q3 = $2100(T_{3}- 10)$
According to the law of heat exchange.
Heat lost by stream = heat gained by water
Q1 + Q2 = Q3
$11300 + 21 (100 – T_{3})$ = $2100 (T_{3}- 10)$
$11300 + 2100 - 21T_{3}$ = $2100 T_{3}$ – 21000
$13400 + 21000 – 21 T_{3}$ = $2100 T_{3} – 21 T_{3}$
34400 = $2121 T_{3}$
$T_{3}$= $\frac{34400}{2121}$
$T_{3}$= 16.2 °C

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