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If a ball of \[80kg\] mass hits an ice cube and the temperature of the ball is \[{100^0}C\] , then how much ice is converted into water ? ( Specific heat of ball is $0.2cal{g^{ - 1}}$ ,latent heat of ice $ = 80ca{\lg ^{ - 1}}$ )
A. $20g$
B. \[200g\]
C. \[2 \times {10^3}g\]
D. \[2 \times {10^4}g\]

Answer
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Hint: We can solve this problem with the help of the concept of thermodynamics. We can calculate the heat released and heat absorbed by material with equation ,$Q = Ms\Delta T$ where $M$ is mass of substance, $s$ is specific heat and $\Delta T$ is the temperature difference or $\Delta T = {T_{final}} - {T_{initial}}$.

Complete step-by-step answer:
The latent heat can be defined as the amount of heat required to change the phase of any material of mass $m$. This can be given by the equation $q = mL$ where $L$ is latent heat. Latent heat is of two types depending upon the change of phase of the material such as latent heat of fusion and latent heat of vaporization.
Let the mass of melted ice is $m$, mass of the ball is $M$, specific heat is $s$, $\Delta T$ is the temperature difference and $L$ is latent heat of ice then we have studied that the heat spend in melting of ice is equal to the heat supplied by the ball. Now;
$
  mL = sM\Delta T \\
  m \times 80 = 0.2 \times (80 \times 1000) \times 100 \\
  m = 2 \times {10^4}g \\
$
Hence \[2 \times {10^4}g\] ice is converted into water. Thus option D is correct answer of this question that is If a ball of \[80kg\] mass hits an ice cube and temperature of ball is \[{100^0}C\], then ice converted into water is \[2 \times {10^4}g\].

So, the correct answer is “Option D”.

Note: We can solve such problems with the help of laws and concepts of thermodynamics. In this problem mass of ball specific heat and latent heat of ice were given in the problem statement. We have put all the values in the equation which was formed by equating the heat spent in melting of ice to the heat supplied by the ball. As all the quantities were known except the mass of melted ice so we get our answer in this way.