Thermology,calorimetry , help with lesson

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The discussion revolves around a thermodynamics problem involving the heat exchange between ice and water in an adiabatic container. The calculations show that 200 g of ice at -20 °C absorbs heat to reach 0 °C, then melts, and finally warms up to the equilibrium temperature of 40 °C. The total heat absorbed by the ice is calculated to be 26,000 calories, leading to the conclusion that the initial mass of water in the container is 650 g. The responses confirm that the algebra and approach used in the calculations are correct. The participant expresses relief after receiving confirmation of their work.
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200 g of ice at -20 ° C are placed in an adiabatic container walls containing certain amount of water at 80 ° C. After heat exchange, temperature equilibrium is θ = 40 ° C. Calculate the quantity of water contained in the container prior to blending. Data: specific heat of ice cG = 0.5 cal / g. ° C, latent heat of fusion of ice L = 80 cal / g, specific heat of water and water = 1 cal / g. ° C

This was my response

M = Q*c * ∆T = 200 * 0.5 * 20 (because it goes until it reaches 0 ° C) = 2000 cal, Q = m * l 200 * 80 = 16000 cal, now the ice is in the liquid state 200*1* 40 =8000 => Qtotal ice as there is heat exchange quantities of heat are balanced => 26000=Ma*1*40 = 650g

right? I'm helping a friend and want to make sure it's right
sorry I do not speak English well, i am from Brazil and i am still in high school . Sorry the inconvenience
 
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Looks good. That's what you're supposed to do, and all the algebra seems to be correct.
 
Thank you. Now I am relaxed
 
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