Three liquids with masses $m_1,m_2,m_3$ are thoroughly mixed. If their specific heats are $c_1,c_2,c_3$ and their temperatures $T_1,T_2,T_3$ respectively, then the temperature of the mixture is
$\frac{{{c_1}{T_1}\, +\, {c_2}{T_2} \,+\, {c_3}{T_3}}}{{{m_1}{c_1}\, +\, {m_2}{c_2} \,+\, {m_3}{c_3}}}$
$\frac{{{m_1}{c_1}{T_1}\, +\, {m_2}{c_2}{T_2} \,+ \,{m_3}{c_3}{T_3}}}{{{m_1}{c_1} \,+\, {m_2}{c_2} \,+\, {m_3}{c_3}}}$
$\frac{{{m_1}{c_1}{T_1} \,+\, {m_2}{c_2}{T_2} \,+\, {m_3}{c_3}{T_3}}}{{{m_1}{T_1} \,+\, {m_2}{T_2} \,+\, {m_3}{T_3}}}$
$\frac{{{m_1}{T_1} \,+\, {m_2}{T_2} \,+\, {m_3}{T_3}}}{{{c_1}{T_1} \,+ \,{c_2}{T_2} \,+\, {c_3}{T_3}}}$
$540\ g$ of ice at $0\ ^oC$ is mixed with $540\ g$ water at $80\ ^oC$. The final temperature of the mixture is
A pendulum clock keeps correct time at $0\,^oC$. The thermal coefficient of linear expansion of the material of the pendulum is $\alpha $. If the temperature rises to $t\,^oC$, then the clock loses per day by (in second)
Two substances $A$ and $B$ of equal mass $m$ are heated at uniform rate of $6\,cal\,s^{-1}$ under similar conditions. A graph between temperature and time is shown in figure. Ratio of heat absorbed $H_A/H_B$ by them for complete fusion is
A lead bullet at $27\ ^oC$ just melts when stooped by an obstacle. Assuming that $25\%$ of heat is absorbed by the obstacle, then the velocity of the bullet at the time of striking ....... $m/s$ ( $M.P.$ of lead $= 327\,^oC$ , specific heat of lead $= 0.03\,cal/g\,^oC$ , latent heat of fusion of lead $= 6\,cal/g$ and $J = 4.2\,joule/cal$ )
The amount of heat required to change $1\ gm (0^o C)$ of ice into water of $100^o C$, is ............ $\mathrm{cal}$