The indian ocean is the third largest ocean at 68,556,000 sq km
Answer:
Tt = 70 + 135e^-0.031t
13 minutes
Explanation:
Given that :
Initial temperature, Ti = 205°
Temperature after 2.5 minutes = 195°
Temperature of room, Ts= 70
Using the relation :
Tt = Ts + Ce^-kt
Temperature after time, t
When freshly poured, t = 0
205 = 70 + Ce^-0k
205 = 70 + C
C = 205 - 70 = 135°
T after 2.5 minutes to find proportionality constant, k
Tt = Ts + Ce^-kt
195 = 70 + 135e^-2.5k
125 = 135e^-2.5k
125 / 135 = e^-2.5k
0.9259 = e^-2.5k
Take In of both sides :
−0.076989 = - 2.5k
k = −0.076989 / - 2.5
k = 0.031
Equation becomes :
Tt = 70 + 135e^-0.031t
t when Tt = 160
160 = 70 + 135e^-0.031k
90 = 135e^-0.031t
90/135 = e^-0.031t
0.6667 = e^-0.031t
In(0.6667) = - 0.031t
−0.405465 = - 0.031t
t = 0.405465/ 0.031
t = 13.071
t = 13 minutes
Answer:
2.55 × 10³ J =2.55 kJ
Explanation:
Specific heat capacity of ice = 37.8 J / mol °C
Specific heat capacity of water = 76.0 J/ mol °C
Ice at -12 °C is converted to ice at 0 °C by absorbing heat Q₁
Ice at 0°C melts to water at 0 °C. Let Heat absorbed during this phase change be Q₂ .
Let heat absorbed to raise the temperature of water from 0 C to 24°C be Q₃ .
Total heat = Q = Q₁ + Q₂ + Q₃
Q₁ = (37.8 j/mol C )(5.53 g /18.01532 g/ mol )( 0-(-12)) = 139.23749 j
Q₂ =(5.53 g/18.01532 g H₂O / mol ) (6.02 x10³ j) = 1847.905 j
Q₃ = (76 j/mol C) ( (5.53 g/18.01532 g H₂O / mol )(24-0) = 559.8968 j
Total Heat required = Q = 139.23749 j + 1847.905 j + 559.8968 j
= 2547.039 j = 2.55 × 10³ J =2.55 kJ
Answer:
Efficiency = 80%
Explanation:
Given the following data;
Work output = 240 N
Work Input = 300 N
To find the mechanical efficiency of a machine;
Substituting into the equation, we have;

Efficiency = 80%
Therefore, the mechanical efficiency of the machine is 80 percent.
Answer:
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