We have half a bucket $(6l)$ of water at $20\,^oC$ . If we want water at $40\,^oC$, how much steam at $100\,^oC$ should be added to it ?
$200\,g$
$\frac {200}{9}\,g$
$2\,kg$
$\frac {200}{3}\,g$
Heat required to convert $5\ kg$ ice at $0\ ^oC$ into water at $100\ ^oC$ is
A liquid of mass $M$ and specific heat $S$ is at a temperature $2t$. If another liquid of thermal capacity $1.5$ times, at a temperature of $\frac{t}{3}$ is added to it, the resultant temperature will be
$300 \,gm$ of water at $25^{\circ} C$ is added to $100 \,gm$ of ice at $0^{\circ} C$. The final temperature of the mixture is ........... $^{\circ} C$
A beaker contains $200\, gm$ of water. The heat capacity of the beaker is equal to that of $20\, gm$ of water. The initial temperature of water in the beaker is $20°C.$ If $440\, gm$ of hot water at $92°C$ is poured in it, the final temperature (neglecting radiation loss) will be nearest to........ $^oC$
In the definition of 'calorie' one calorie is the heat required to raise the temperature of $1\ gram$ of water through $1\ ^oC$ in a certain interval of temperature. The temperature interval is