Question

Can you please help me to answer this question? Please provide detailed explanation on how you get to the answer. Please do not omit any steps and please explain in detail specially part b. Thank you very much in advance!

Suppose you have a perfectly insulated container holding 2.00 kg of liquid water at a temperature of 20.0 °C. You also have a large supply of ice cubes which are stored at a temperature of -20.0 °C. a. What is the maximum mass of ice you can add to the water so that all of the ice melts? Assume no heat is lost to the environment. b. Suppose you add twice the mass of ice that you found in part a. How much solid ice will remain once thermal equilibrium is reached?

Answer 1

maximum heat loss by water = m_ws_w(20) = 2 x 4186 x 20 = 167440 J

now let m kg mass of ice canbe melted

then mL + ms_i(20) = 167440

=> m(334000) + m(2109)(20) = 167440

=> m = 0.4451 kg

part b )

absorbed Meat by ice - moss, uf ice x sp. heat change in temp = 2x0.4451 2090 x 20 = 37,210.36 J released by watu = 2x4480 x 20 - 167,200 J Heat Meat is used - to meet the ice 167,200w-37,310.26 = 129, 989.64 J I mars as ice melted x latent hext = 129, 989.647 mass free melted = 129,989.64 = 0.389 4 334 x 103 mass of ice left f. 260.4451) - (0.389) = - 0.56) ke