$A$ solid cube and a solid sphere of identical material and equal masses are heated to the same temperature and left to cool in the same surroundings. Then,

  • A
    the cube will cool faster because of its sharp edges
  • B
    the cube will cool faster because it has a larger surface area
  • C
    the sphere will cool faster because it is smooth
  • D
    the sphere will cool faster because it has a larger surface area

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The temperature of a body in air falls from $40^{\circ} \text{C}$ to $24^{\circ} \text{C}$ in $4 \text{ minutes}$. The temperature of the air is $16^{\circ} \text{C}$. The temperature of the body in the next $4 \text{ minutes}$ will be $:$

$A$ bucket full of hot water cools from $75^{\circ}C$ to $70^{\circ}C$ in time $T_1$,from $70^{\circ}C$ to $65^{\circ}C$ in time $T_2$,and from $65^{\circ}C$ to $60^{\circ}C$ in time $T_3$. Which of the following relations is correct?

$A$ calorimeter contains $10 \ g$ of water at $20^{\circ}C$. The temperature falls to $15^{\circ}C$ in $10 \ min$. When the calorimeter contains $20 \ g$ of water at $20^{\circ}C$,it takes $15 \ min$ for the temperature to become $15^{\circ}C$. The water equivalent of the calorimeter is: (in $g$)

$A$ hot body is allowed to cool. The surrounding temperature is constant at $30^{\circ} C$. It takes time $t_{1}$ to cool from $70^{\circ} C$ to $68^{\circ} C$ and time $t_{2}$ to cool from $60^{\circ} C$ to $59.5^{\circ} C$. Then:

$A$ body cools from $70^{\circ} C$ to $40^{\circ} C$ in $5 \text{ min}$. Calculate the time it takes to cool from $60^{\circ} C$ to $40^{\circ} C$. The temperature of the surrounding is $20^{\circ} C$. (in $\text{ min}$)

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