FPIsaac
- 4
- 0
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
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