Find the result of mixing 10g of ice at -10 celsius with 10g of water at 10 celsius. Specific heat capacity of ice = 2.1 J/gK,specific latent heat of ice = 336 x1000J/g and and specific heat capacity of water = 4.2 J/gK.??

PS : Sir PLEASE explain each step because I didn't understand all the steps in a solution given for this same sum earlier.PLEASE DONT GIVE A LINK TO A SIMILAR ANSWER .....Thanks

Specific heat of water is = 4.2 J/(g.K)

Specific heat of ice is = 2.1 J/(g.K)

Latent heat of ice is = 336 J/g

Heat released by 10 g of water at 10 ^oC in converting to water at 0 ^oC is,

H₁ = msθ = (10)(4.2)(10) = 420 J

Heat gained by 10 g of ice at -10 ^oC in converting into ice at 0 ^oC is,

H₂ = (10)(2.1)(10) = 210 J

Suppose, ‘x’ g of ice now melts. So, heat gained by this ‘x’ g of ice in converting into water at 0 ^oC is,

H₃ = xL = x336 = 336x J

Now, H₃ + H₂ = H₁

=> 336x + 210 = 420

=> x = 0.625 g

Mass of ice remaining is = 10 – 0.625 = 9.375 g

Thus, in the resulting mixture the ice remaining is 9.375 g and the temperature of the mixture is 0 ^oC.
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