26.0 g steam at 120 °C is added temperature of the resulting water 120 g of...

Question

Question

26.0 g steam at 120 °C is added temperature of the resulting water 120 g of ice at -10 °C in an insulated container. What is

Answer 1

Answer #1

Date: 1201 -3 mass of steam m = 26.0g m = 26.0 g = 26x10 kg Temperature of steam Ti = 12000 T = 393k = 120+273 20 X 163 kg =

Answer 2

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