A drilling machine of $10\,KW$ power is used to drill a bore in a small aluminium block of mass $8\,kg.$ If $50\%$ of power is used up in heating the machine itself or lost to the surroundings then ........ $^oC$ is the rise in temperature of the block in $2.5\,minutes$
[specific heat of aluminium $= 0.91\,J/g\,\,^oc$ ]
$103$
$130$
$105$
$30$
We have half a bucket ($6$ litre) of water at $20^oC $.If we want water at $40^oC$, how much steam at $100^oC$ should be added to it ?
A calorimeter of water equivalent $20\, g$ contains $180\, g$ of water at $25^{\circ} C$. '$m$' grams of steam at $100^{\circ} C$ is mixed in it till the temperature of the mixure is $31^{\circ} C$. The value of $'m'$ is close to
(Latent heat of water $=540$ cal $g ^{-1}$, specific heat of water $=1$ cal $g^{-1}{ }^{\circ} C ^{-1}$ )
$10 \,gm$ of ice at $-20^{\circ} C$ is kept into a calorimeter containing $10 \,gm$ of water at $10^{\circ} C$. The specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain ..........
Calculate the heat required to convert $3\; kg$ of ice at $-12\,^{\circ} C$ kept in a calorimeter to steam at $100\,^{\circ} C$ at atmospheric pressure. Given specific heat capacity of $ice =2100\; J \,kg ^{-1} K ^{-1}$. specific heat capacity of water $=4186\; J kg ^{-1} K ^{-1}$, latent heat of fusion of ice $=3.35 \times 10^{5} \;J \,kg ^{-1}$ and latent heat of steam $=2.256 \times 10^{6}\; J \,kg ^{-1}$
A bullet of mass $10 \,g$ moving with a speed of $20 \,m / s$ hits an ice block of mass $990 \,g$ kept on a frictionless floor and gets stuck in it. How much ice will melt if $50 \%$ of the lost KE goes to ice is .......... $g$ (initial temperature of the ice block and bullet $=0^{\circ} C$ )