# 35 - Heat and Thermodynamics Questions Answers

50gm ice at 0 degree celsius and 100gm water at 20 degree celsius are mixed, find final temperature and composition of the mixture.

**Asked By: TARSH SWARNKAR**

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6 identical rods are arranged in such a way that 3 of them make an equilateral triangle and other 3 are extended at the vertices forming a Y shape at each vertex. Name the triangle ABC , and other ends of rods as P, Q , R . given that temperatures at P , Q and R are 200°c,20°c and 20°c , what are temperatures at A , B , and C?

**Asked By: AHAD QURESHI**

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**Joshi sir comment**

A gas expands against a variable external pressure given by **p = 10/V atm**, where **V **is the volume at each stage of expansion. in expanding from **10 L **to **100 L**, the gas undergoes a change in internal energy **Δ****U = 418 J**. How much heat has been absorbed?

**Asked By: RICHIK BANDYOPADHYAY**

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**Joshi sir comment**

process is isothermal because pV = 10 = nRT

so work done W = 2.303nRTlog(V_{2}/V_{1})

so Q = **Δ****U **+ W

The temperature of a perfect black body

is 1000k and its area

is 0.1m^{2} . the heat radiated by it in 1 minute in joule is

**Asked By: HASITHA.KOTTA**

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**Joshi sir comment**

Q/At = σT^{4}

now solve

A CARNOT ENGINE WORKING B/W 27^{0} C & --123^{0}C HAS EFFICIENCY η . IF TEMP OF SINK IS DECREASED BY 50K THE EFFICIENCY BECOMES??

ANS 4η/3

**Asked By: SARIKA**

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**Joshi sir comment**

according to the given condition η = 1-(150/300) = 0.5

now after decreasing temperature of sink by 50K

we get the new efficiency = 1-(100/300) = 2/3 = 4*0.5/3 = 4η/3

in a thermodynamic process on 2 moles of a monoatomic gas work done on gas is 100J & change in temp is 30 C . The change in internal energy of the gas in this process is??

**Asked By: SARIKA**

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**Joshi sir comment**

change in internal energy = nC_{v}dT = 2*[R/(γ-1)]*30 = 60*8.31/[(5/3)-1]

2 MOLES OF MONOATOMIC GAS IS MIXED WITH 1 MOLE OF A DIATOMIC GAS. THEN γ FOR THE MIXTURE??

ANS 1.55

**Asked By: SARIKA SHARMA**

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**Joshi sir comment**

total no of freedom for this mixture = (2*3+1*5)/(2+1) = 11/3

so γ = 1+[2/f] = 1 + 2/[11/3] = 1 + [6/11] = 17/11 = 1.55

a thin square plate with each side equal to 10cm , is heated by a blacksmith . the rate radiated energy by the heated plate is 1134W the temp of hot square plate is

( σ = 5.67 X 10^{-8} Wm^{2}/k^{4} emissivity of plate =1)

ANS 1000K

**Asked By: SARIKA SHARMA**

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**Joshi sir comment**

according to the Stefans law P = AeσT^{4}

so 1134 = 0.02*1*5.67*10^{-8}*T^{4 }(area of both the faces are considered)

so 1134*10^{10}/11.34 = T^{4}

so T = 1000