Specific heat of alcohol is about half that of water. Suppose you have identical masses of alcohol a

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10-1.Thermometry, Thermal Expansion and Calorimetry

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The specific heat of alcohol is about half that of water. Suppose you have identical masses of alcohol and water. The alcohol is initially at temperature $T_A$ . The water is initially at a different temperature $T_W$ . Now the two fluids are mixed in the same container and allowed to come into thermal equilibrium, with no loss of heat to the surroundings. The final temperature of the mixture will be

closer to $T_A$ than $T_W$

closer to $T_W$ than $T_A$

exactly halfway between $T_A$ and $T_W$

dependent on the volume of alcohol used.

Solution

Final common temperature

$\mathrm{T}=\frac{\mathrm{m}_{1} \mathrm{S}_{1} \mathrm{T}_{1}+\mathrm{m}_{2} \mathrm{S}_{2} \mathrm{T}_{2}}{\mathrm{m}_{1} \mathrm{S}_{1}+\mathrm{m}_{2} \mathrm{S}_{2}}$

because $\mathrm{m}_{1}=\mathrm{m}_{2}$ given

$\therefore \quad \mathrm{T}=\frac{\mathrm{S}_{\mathrm{A}} \mathrm{T}_{\mathrm{A}}+\mathrm{S}_{\mathrm{w}} \mathrm{T}_{\mathrm{w}}}{\mathrm{S}_{\mathrm{A}}+\mathrm{S}_{\mathrm{w}}}$

$\& \mathrm{S}_{\mathrm{A}}=\frac{1}{2} \mathrm{S}_{\mathrm{w}}$ given

$\mathrm{T}=\frac{\frac{\mathrm{T}_{\mathrm{A}}}{2}+\mathrm{T}_{\mathrm{w}}}{\frac{1}{2}+1}=\left(\frac{\mathrm{T}_{\mathrm{A}}+2 \mathrm{T}_{\mathrm{w}}}{3}\right)$

$\mathrm{T}=\left(\frac{\mathrm{T}_{\mathrm{A}}}{3}+\frac{2 \mathrm{T}_{\mathrm{w}}}{3}\right) \&$ This $\mathrm{T}$ is nearer to $\mathrm{T}_{\mathrm{w}}$

$Method-II:$

$\mathrm{S}_{\mathrm{A}}=\frac{1}{2} \mathrm{S}_{\mathrm{w}}$ given

so, water has more thermal inertia, hence in mixture, it will loose less temperature $\&$ final temperature will be nearer to $\mathrm{T}_\mathrm{w'}$

Standard 11

Physics

Pure water super cooled to $-15^o C$ is contained in a thermally insulated flask. Small amount of ice is thrown into the flask. The fraction of water frozen into ice is :

hard

An electric kettle (rated accurately at $2.5\, kW$) is used to heat $3\, kg$ of water from $15\,^oC$ to boiling point. It takes $9.5$ minute. Then, the amount of heat that has been lost is

medium

Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature $T_0$, while Box contains one mole of helium at temperature $(7/3)$ $T_0$ The boxes are then put into thermal contact with each other, and heat flows between them until the gases reach a common final temperature (ignore the heat capacity of boxes). Then, the final temperature of the gases,$T_f$ in terms of $T_0$ is

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(AIEEE-2006)

When $x\, grams$ of steam at $100\,^oC$ is mixed with $y\,grams$ of ice at $0\,^oC$ , We obtain $(x + y)\,grams$ of water at $100\,^oC$ . What is the ratio $y/x$ ?

hard

$1\ gm$ of ice at $0^o C$ is mixed with $1gm$ of water at $100^o C$ the resulting temperature will be ………. $^oC$

medium

(AIIMS-1994)

Solution

Similar Questions

Pure water super cooled to $-15^o C$ is contained in a thermally insulated flask. Small amount of ice is thrown into the flask. The fraction of water frozen into ice is :

An electric kettle (rated accurately at $2.5\, kW$) is used to heat $3\, kg$ of water from $15\,^oC$ to boiling point. It takes $9.5$ minute. Then, the amount of heat that has been lost is

When $x\, grams$ of steam at $100\,^oC$ is mixed with $y\,grams$ of ice at $0\,^oC$ , We obtain $(x + y)\,grams$ of water at $100\,^oC$ . What is the ratio $y/x$ ?

$1\ gm$ of ice at $0^o C$ is mixed with $1gm$ of water at $100^o C$ the resulting temperature will be ………. $^oC$

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