Thermal Properties Of Matter

Q: A steel scale is correct at 0^o C , the length of a brass tube measured by it at 40^o C is 5 m . The correct length of the tube at 0^o C is (Coefficients of linear expansion of steel and brass are 11 \times 10^{-6/^o C } and 19 \times 10^{-6/^o C } respectively).

l_s'=l_b'=5 m T_f=40^0c Actual length of brass tube, l_b=? T_i=0 For Length of scale at 0^0=l_s, l_s'=l_s(1+\alpha_s \Delta T) 5=l_s(1+40\alpha_s) l_s=\dfrac{5}{1+40\alpha_s} Actual length of tube at 40^0 l_b'=l_s'+\Delta l_s =l_s'+l'_s-l_s =2l_s'-l_s l_b'=\left(2\times 5-\dfrac{5}{1+40\alpha_s}\right) Actual length of tube at 0^0c=l_b l_b'=l_b(1+\alpha_0 4T) l_b=\dfrac{l_b'}{1+\alpha_b4T}=(10-\dfrac{5}{1+40\alpha_s})\times \dfrac{1}{1+40\alpha_b} =\left(10-\dfrac{5}{1+40\alpha_s}\right)\times \dfrac{1}{1+40\alpha_b} =\left(10-\dfrac{5}{1+40\times 11\times 10^{-6}}\right)\times \left(\dfrac{1}{1+40\times 19\times 10^{-6}}\right) =\left(10-\dfrac{5}{1+440\times 10^{-6}}\right)\times \left(\dfrac{1}{1+760\times 10^{-6}}\right) =\left(10-\dfrac{5}{1.000440}\right)\times \left(\dfrac{1}{1+00076}\right) =4.998\cong 4.999m \text {or} Let the length of tube and scale at 0^o C be l. Then,Change in measured tube length \Delta l= Thermal expansion of brass - Thermal expansion of steel \Delta l=l\alpha_B\Delta T-l\Delta\alpha_B\Delta T \Delta l=l\Delta T(\alpha B-\alpha S) \Delta l=l(40-0)(19\times 10^{-6}-11\times 10^{-6}) So, new length measured by scale =l+\Delta l \Rightarrow 5=l+l(40)(7\times 10^{-6}) \Rightarrow \dfrac{5}{1+40\times 7\times 10^{-6}}=l \Rightarrow l=4.9986\approx 4.999 mOption C is correct.

Q: 2 kg of ice at -20 ^o C is mixed with 5 kg of H _2 O at 20^o C in an insulating vessel having negligible heat capacity. Calculate the final mass of water left in the container.Given: Specific heats of water and ice are 1 kcal kg ^{-1} ^o C ^{-1} and 0.5 kcal kg ^{-1 o} C ^{-1} and latent heat of fusion of ice is 80 k cal kg ^{-1} .

M_i = 2kg T_i = -20^oC M_w = 5kg T_w = 20^oC Assume all ice and water come in thermal equilibrium at 0^oC ice. H = -H_{(ice - 20 to 0)} + Water_{(20\to 0)} + Water_{(0^o ice)} =-m_is_i \Delta T + M_wS_w \Delta T + M_w L =-2\times 5\times 20 + .5 \times 1 \times 20 + 5 \times 80 = -20 + 100 + 400 H = +480 Heat come positive, so this heat will melt the ice from 0^o ice to 0^oC waterNow, total mass of ice = 2+5=7kg .Must that convert into water from given heat H = 480 H = M'L M' = \dfrac{H}{L} = \dfrac{480}{80} M' =6kg Mass of water =6kg ice =1kg

Q: Three rods made of the same material and having the same cross-section have joined as shown in the fig. Each rod is of same length. The left and right ends are kept at 10^0 C and 90^0 C respectively. The temperature of the junction of the three rods will be :

Let the temp at the intersection of the three rods is T .Since at steady state rate of heat transfer is same, KA\dfrac{(90-T)}{x} + KA\dfrac{(90-T)}{x} = KA\dfrac{(T-10)}{x} \Rightarrow 2(90-T)= T-10 \Rightarrow T={\dfrac{190}{3}}^0C

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Problem solving tips

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- Want to score better marks in exams? Have a look at some problem solving tips.

Tips: Calculate actual measurement with scale at different temperature
In the scenario when the length of scale and object changing with temperature and we have to find correct measurement using the given scale. Use following strategy :
Let Scale gives correct measurement temperature

T

and we have given scale measurement at

T^{'}

(say

l^{'}

).
Step 1: Find length of scale at temperature

T

(Assume

l

).
Step 2: Find change in length of scale

Δ l

.
So, Actual measurement of object at temperature

T^{'}

would be

l^{'} \pm Δ l = l^{'} \pm (l^{'} - l)

Now we know the actual length of given object at temperature

T^{'}

.
Using above measurement of object we will find the length of object at all given temperature

A steel scale is correct at

0^{o} C

, the length of a brass tube measured by it at

4 0^{o} C

5 m

. The correct length of the tube at

0^{o} C

is (Coefficients of linear expansion of steel and brass are

11 \times 1 0^{- 6 /^{o} C}

and

19 \times 1 0^{- 6 /^{o} C}

respectively).

4.001 m

5.001 m

4.999 m

4.501 m

View solution

Tips 2: Tips to Calculation of ratio of ice and water present at equilibrium temperature
If in a question you have given that some amount of ice is added to limited amount of water at certain temperature. From the question, if you are not sure what will be the final state of a mixture (water or ice) then follow the following steps

Assume, the heat released is positive heat, Heat absorbed is negative heat.
Calculate heat involved to bring all ice and water at $0^{\circ} C$ ice. (After doing this Container have only ice at $0^{\circ} C$ )
- If value of heat is coming positive: Ice will melt according to amount of positive heat.
- If all ice melt into water and still heat is in excess amount then temperature of water will increase.
- If heat is not sufficient to melt all ice then we will see water and ice in the container.
If value of heat is showing negative: it mean that at equilibrium ice will exist and its temperature will be below $0^{\circ} C$

2

kg of ice at

- 20

^{o}

C is mixed with

5

kg of H

_{2}

O at

2 0^{o}

C in an insulating vessel having negligible heat capacity. Calculate the final mass of water left in the container.
Given: Specific heats of water and ice are

1

kcal kg

^{- 1}

^{o}

^{- 1}

and

0.5

kcal kg

^{- 1 o}

^{- 1}

and latent heat of fusion of ice is

80

k cal kg

^{- 1}

7 kg

6 kg

4 kg

2 kg

View solution

Tips 3: Tips to calculate temperature at the junction of multiple conductor
In the questions, if you have given the end temperature of every conductor and you have to find temperature at the junction then follow following tip.

Assume temperature of junction is $T$ .
In given steady condition, Assume heat is moving from junction to outer end of every conductor. So sum of heat current at the junction would be zero.

ie;

i_{1} + i_{2} + i_{3} + i_{4} = 0

\frac{T - T _{1}}{R _{1}} + \frac{T - T _{2}}{R _{2}} + \frac{T - T _{3}}{R _{3}} + \frac{T - T _{4}}{R _{4}} = 0

Where R: Represent thermal resistance of respective conductors.

Three rods made of the same material and having the same cross-section have joined as shown in the fig. Each rod is of same length. The left and right ends are kept at

1 0^{0}

C and

9 0^{0}

C respectively. The temperature of the junction of the three rods will be :

7 0^{0} C

6 0^{0}

3 0^{0}

\frac{1 9 0}{3}^{0} C

View solution