540, g of ice at 0^oC is mixed with 540, g of water at 80^oC . final temperature of mixture is

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10-1.Thermometry, Thermal Expansion and Calorimetry

hard

$540\, g$ of ice at $0^oC$ is mixed with $540\, g$ of water at $80^oC$. The final temperature of the mixture is

A

$0°C$

B

$40°C$

C

$80°C$

D

Less than $0°C$

Solution

(a) Heat taken by ice to melt at $0°C$ is

${Q_1} = mL = 540 \times 80 = 43200\,cal$

Heat given by water to cool upto $0°C$ is

${Q_2} = ms\Delta \theta = 540 \times 1 \times (80 – 0) = 43200\,cal$

Hence heat given by water is just sufficient to melt the whole ice and final temperature of mixture is $0°C.$

Short trick : For these type of frequently asked questions you can remember the following formula

${\theta _{{\rm{mix}}}} = \frac{{{m_W}{\theta _W} – \frac{{{m_i}{L_i}}}{{{c_W}}}}}{{{m_i} + {m_W}}}$ (See theory for more details)

If ${m_W} = {m_i}$ then ${\theta _{mix}} = \frac{{{\theta _W} – \frac{{{L_i}}}{{{c_W}}}}}{2} = \frac{{80 – \frac{{80}}{1}}}{2} = 0^\circ C$

Standard 11

Physics

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