$200 \, g$ of a solid ball at $20 \, ^oC$ is dropped into an equal amount of water at $80 \, ^oC$. The resulting temperature is $60 \, ^oC$. This means that the specific heat of the solid is

  • A
    One fourth of water
  • B
    One half of water
  • C
    Twice of water
  • D
    Four times of water

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Similar Questions

In a calorimeter of water equivalent $20 \,g$,water of mass $1.1 \,kg$ is taken at $288 \,K$ temperature. If steam at temperature $373 \,K$ is passed through it and the temperature of the water increases by $6.5^{\circ} C$,then the mass of steam condensed is ............ $g$.

How many grams of ice at $0 \, ^\circ \text{C}$ will be melted by $1 \, \text{g}$ of steam at $100 \, ^\circ \text{C}$? (Latent heat of fusion of ice $L = 80 \, \text{cal/g}$ and latent heat of vaporization of water $L' = 540 \, \text{cal/g}$)

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Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is held at a temperature of $100^{\circ} C$,while the other one is kept at $0^{\circ} C$. If the two are brought into contact,then assuming no heat loss to the environment,the final temperature that they will reach is

$M$ grams of steam at $100^{\circ} C$ is mixed with $200 \; g$ of ice at its melting point in a thermally insulated container. If it produces liquid water at $40^{\circ} C$ [heat of vaporization of water is $540 \; cal/g$ and heat of fusion of ice is $80 \; cal/g$],the value of $M$ is:

$0.1 \,m^{3}$ of water at $80^{\circ} C$ is mixed with $0.3 \,m^{3}$ of water at $60^{\circ} C$. The final temperature of the mixture is (in $^{\circ} C$)

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