Answer:
Figure attached
We can conclude that majority of the values are positive. And we can say that is skewed to the right because the Median< Mean is and we have most of the values at the left of the distribution.
Explanation:
We can use the following R code to obtain the data for wdiff:
source("http://www.openintro.org/stat/data/cdc.R") #obtain the info
nrow(cdc) # number of elements
names(cdc) # obtain the name for the variable
[1] "genhlth" "exerany" "hlthplan" "smoke100" "height" "weight" "wtdesire" "age"
[9] "gender"
wdiff represent the difference between desired weight (wtdesire) and current weight (weight) and we can obtain the data with the following code:
wdiff <- (cdc$weight-cdc$wtdesire)
And now we can create the histogram with this code
hist(wdiff,xlim =c(-100,150))
> mean(wdiff)
[1] 14.5891
> median(wdiff)
[1] 10
And the result is on the figure attached.
And we can conclude that majority of the values are positive. And we can say that is skewed to the right because the Median< Mean is and we have most of the values at the left of the distribution.
To solve this problem it is necessary to apply the concepts related to the heat flux rate expressed in energetic terms. The rate of heat flow is the amount of heat that is transferred per unit of time in some material. Mathematically it can be expressed as:
Where
k = 0.84 J/s⋅m⋅°C (The thermal conductivity of the material)
Area
Length
= Temperature of the "hot"reservoir
= Temperature of the "cold"reservoir
Replacing with our values we have that,
Therefore the correct answer is B.
Answer : The rate of heat transfer to the water is, 37.92 kJ/min
Explanation : Given,
Time = 10 min
Mass of water = 200 g
Latent heat of fusion of water = 334 J/g
Latent heat of vaporization of water = 2230 J/g
Now we have to calculate the rate of heat transfer to the water.
Now put all the given values in the above formula, we get:
Thus, the rate of heat transfer to the water is, 37.92 kJ/min
Answer:
3349J/kgC
Explanation:
Questions like these are properly handled having this fact in mind;
Quantity of heat = mcΔ∅
m = mass of subatance
c = specific heat capacity
Δ∅ = change in temperature
m₁c₁(∅₂-∅₁) = m₂c₂(∅₁-∅₃)
m₁ = mass of block = 500g = 0.5kg
c₁ = specific heat capacity of unknown substance
∅₂ = block initial temperature = 50oC
∅₁ = equilibrium temperature of block and water after mix= 25oC
m₂= mass of water = 2kg
c₂ = specific heat capacity of water = 4186J/kg C
∅₃ = intial temperature of water = 20oC
0.5c₁(50-25) = 2 x 4186(25-20)
And we can find c₁ which is the unknown specific heat capacity
c₁ = = 3348.8J/kg C≅ 3349J/kg C