Suppose 25 gm of steam at 120°C is mixed with 25 g of ice at -30°C. What will be the final temperature?
Answers
Answered by
12
Answer:
-25 oc
here is the correct answer please mark me brainliest and give thanks
Answered by
0
Explanation:
we know that
total heat loss(steam) = total heat gain(ice)
25×0.48(120-t) = 25× 0.5(t-(-30))
0.48(120-t) = 1/2(t+30)
0.96(120-t) = t+30
115.2-0.96t = t+30
115.2 - 30=t + 0.96t
85.2 = 1.96t
t = 43.46°C
hope it helps...
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