What is the rate of heat transfer required to melt 1-ton of ice at 32 F in 24 hours?
1 answer:
Answer:
3865.74 J/s
Step-by-step explanation:
mass of ice, m = 1 ton = 1000 kg
time , t = 24 hours
latent heat of fusion of ice, L = 334000 J/kg
Heat required to melt, H = m x L
where, m is the mass of ice and L be the latent heat of fusion
So, H = 1000 x 334000 = 334 xx 10^6 J
Rate of heat transfer = heat / time =
Rate of heat transfer = 3865.74 J/s
thus, the rate of heat transfer is 3865.74 J/s.
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Step-by-step explanation:
In a two tailed test the probability of occurrence is the total area under the critical range of values on both the sides of the curve (negative side and positive side)
Thus, the probability values for a two tailed test as compared to a one tailed test is given by the under given relation -
\
Here
Substituting the given value in above equation, we get -
probability values for a two tailed test
=