Answer:
(a) 209 Watt
(b) 4482.8 seconds
Explanation:
(a) P = 50×4.18
Where P = rate of heat loss in watt
P = 209 Watt
Applying,
Q = cm(t₁-t₂)................ Equation 1
Where Q = amount of heat given off, c = specific heat capacity capacity of human, m = mass of the person, t₁ and t₂ = initial and final temperature.
From the question,
Given: m = 90 kg, t₁ = 40°C, t₂ = 37°C
Constant: c = 3470 J/kg.K
Substtut these values into equation 1
Q = 90×3470(40-37)
Q = 936900 J
But,
P = Q/t.............. Equation 2
Where t = time
t = Q/P............ Equation 3
Given: P = 209 Watt, Q = 936900
Substitute into equation 3
t = 936900/209
t = 4482.8 seconds
Answer:
209.68 years
Explanation:
Let T be the half life.
t = 1 year
N = 99.67 % of No = 0.9967 No
Use the law of radioactivity
N = No x e ^(- λ t)
Where, λ is decay constant.
λ = 0.6931 / T
So,
0.9967 No = No x e^(- λ t)
0.9967 = e^(- λ t)
e^( λ t) = 1 / 0.9967 = 1.0033
λ t = 3.3 x 10^-3
(0.6931 x 1) / T = 3.3 x 10^-3
T = 209.68 years
Thus, the halflife is 209.68 years.
Answer:
4 %
2 ) 3.42 %
Explanation:
Sensitivity requirement of 4 milligram means it is not sensitive below 4 milligram or can not measure below 4 milligram .
Given , 4 milligram is the maximum error possible .
Measured weight = 100 milligram
So percentage maximum potential error
= (4 / 100) x 100
4 %
2 )
As per measurement
weight of 6 milliliters of water
= 48.540 - 42.745 = 5.795 gram
6 milliliters of water should measure 6 grams
Deviation = 6 - 5.795 = - 0.205 gram.
Percentage of error =(.205 / 6 )x 100
= 3.42 %
