An object kept in a large room having an air temperature of \(25^\circ \text{C}\) takes \(12 ~\text{min}\) to cool from \(80^\circ \text{C}\) to \(70^\circ \text{C}.\) The time taken to cool for the same object from \(70^\circ \text{C}\) to \(60^\circ \text{C}\) would be nearly:
1. \(10 ~\text{min}\)
2. \(12 ~\text{min}\)
3. \(20 ~\text{min}\)
4. \(15 ~\text{min}\)

A deep rectangular pond of surface area \(A\), containing water (density = \(\rho,\) specific heat capacity = \(s\)), is located in a region where the outside air temperature is at a steady value of \(-26^{\circ}\text{C}\)$$. The thickness of the ice layer in this pond at a certain instant is \(x\). Taking the thermal conductivity of ice as \(k\), and its specific latent heat of fusion as \(L\), the rate of increase of the thickness of the ice layer, at this instant, would be given by:
1. \(\dfrac{26k}{x\rho L-4s}\)
2. \(\dfrac{26k}{x^2\rho L}\)
3. \(\dfrac{26k}{x\rho L}\)
4. \(\dfrac{26k}{x\rho L+4s}\)

Two holes are cut into a metal sheet. The diameters of the two holes are \(d_1\) and \(d_2\) \((d_1>d_2).\) If the temperature of the metal sheet is increased, then which of the following distance increases?

In the Arctic region, hemispherical houses called Igloos are made of ice. It is possible to maintain a temperature inside an Igloo as high as \(20^\circ \text{C}\) because:

In a steel factory, it is found that to maintain \(M\) kg of iron in the molten state at its melting point, an input power \(P\) watt is required. When the power source is turned off, the sample completely solidifies in time \(t\) seconds. The latent heat of the fusion of iron is:

Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod has the same length. The left and right ends are kept at \(0^{\circ}\text{C}~\text{and}~90^{\circ}\text{C},\) respectively. The temperature at the junction of the three rods will be:

           
1. \(45^{\circ}\text{C}\)
2. \(60^{\circ}\text{C}\)
3. \(30^{\circ}\text{C}\)
4. \(20^{\circ}\text{C}\)

Hot coffee in a mug cools from \(90^{\circ}\text{C}\) to \(70^{\circ}\text{C}\) in \(4.8\) minutes. The room temperature is \(20^{\circ}\text{C}.\) Applying Newton's law of cooling, the time needed to cool it further by \(10^{\circ}\text{C}\) should be nearly:

One kilogram of ice at \(0^\circ \text{C}\) is mixed with one kilogram of water at \(80^\circ \text{C}.\) The final temperature of the mixture will be: (Take: Specific heat of water = \(4200~\text{J kg}^{-1}\text{K}^{-1},\) latent heat of ice\(=336~\text{kJ kg}^{-1}\))

The value of the coefficient of volume expansion of glycerin is \(5\times10^{-4}\) K^-1. The fractional change in the density of glycerin for a temperature increase of \(40^\circ \mathrm{C}\) will be:

The temperature of a body falls from \(50^{\circ}\text{C}\)$\mathrm{}$ to \(40^{\circ}\text{C}\)$\mathrm{}$ in \(10\) minutes. If the temperature of the surroundings is \(20^{\circ}\text{C},\)$\mathrm{ t}$hen the temperature of the body after another \(10\) minutes will be:
1. \(36.6^{\circ}\text{C}\)          
2. \(33.3^{\circ}\text{C}\)$\mathrm{}$
3. \(35^{\circ}\text{C}\)             
4. \(30^{\circ}\text{C}\)$\mathrm{}$