$A$ thermos flask contains $250 \ g$ of coffee at $90^{\circ} C$. To this $20 \ g$ of milk at $5^{\circ} C$ is added. After equilibrium is established, the temperature of the liquid is (Assume no heat loss to the thermos bottle. Take specific heat of coffee and milk as $1.00 \ cal/g^{\circ} C$) (in $^{\circ} C$)

  • A
    $3.23$
  • B
    $3.15$
  • C
    $83.7$
  • D
    $37.8$

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Similar Questions

$A$ water heater of power $2000\,W$ is used to heat water. The specific heat capacity of water is $4200\,J\,kg^{-1}\,K^{-1}$. The efficiency of the heater is $70\%$. The time required to heat $2\,kg$ of water from $10^{\circ}C$ to $60^{\circ}C$ is $..........s$. (Assume that the specific heat capacity of water remains constant over the temperature range).

When $M_1$ gram of ice at $-10\,^{\circ}C$ (specific heat $= 0.5\, cal\, g^{-1}\,^{\circ}C^{-1}$) is added to $M_2$ gram of water at $50\,^{\circ}C$,finally no ice is left and the water is at $0\,^{\circ}C$. The value of latent heat of ice,in $cal\, g^{-1}$ is

$A$ piece of metal weighing $100 \,g$ is heated to $80^{\circ} C$ and dropped into $1 \,kg$ of cold water in an insulated container at $15^{\circ} C$. If the final temperature of the water in the container is $15.69^{\circ} C$,the specific heat of the metal in $J / g^{\circ} C$ is:

$37 \ g$ of ice at $0^{\circ} C$ temperature is mixed with $74 \ g$ of water at $70^{\circ} C$ temperature. The resultant temperature is (Specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$ and latent heat of fusion of ice $= 80 \ cal \ g^{-1}$) (in $^{\circ} C$)

$A$ liquid of specific heat $0.8 \ cal / g^{\circ} C$ at temperature $60^{\circ} C$ is mixed with another liquid of the same mass having temperature $45^{\circ} C$. If the temperature of the mixture is $53^{\circ} C$,then the specific heat (in $cal / g^{\circ} C$) of the second liquid is:

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