1. Tardigrade
  2. Question
  3. Physics
  4. Four cubes of ice at -10° C each one gm is taken out from the refrigerator and are put in 150 gm of water at 20° C. The temperature of water when thermal equilibrium is attained. Assume that no heat is lost to the outside and water equivalent of container is 46 gm. (Specific heat capacity of water =1 cal / gm - ° C, Specific heat capacity of ice =0.5 cal / gm - ° C, Latent heat of fusion of ice =80 cal / gm - ° C )

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Q. Four cubes of ice at $-10^{\circ} C$ each one $gm$ is taken out from the refrigerator and are put in $150\, gm$ of water at $20^{\circ} C$. The temperature of water when thermal equilibrium is attained. Assume that no heat is lost to the outside and water equivalent of container is $46\, gm$. (Specific heat capacity of water $=1\, cal / gm -{ }^{\circ} C$, Specific heat capacity of ice $=0.5\, cal / \,gm -{ }^{\circ} C$, Latent heat of fusion of ice $=80\, cal / gm -{ }^{\circ} C$ )

Thermal Properties of Matter

Solution:

Heat gained by ice $=$ Heat lost by water $+$ Heat lost by container
Initial temperature of container $=20^{\circ} C$
$4 \times \frac{1}{2} \times 10 +4 \times 80+4 \times 1 x(T-0)$
$=196 \times 1 \times(20-T)$
$20+320+4 T=196 \times 20-196 T$
$200 T=196 \times 20-340 T$
$=\frac{3580}{200}=17.9^{\circ} C$