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algol [13]
2 years ago
10

How much heat is required to raise the temperature of

Mathematics
1 answer:
Elina [12.6K]2 years ago
4 0

Answer:

The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .

Step-by-step explanation:

Given as :

The mass of liquid water = 50 g

The initial temperature = T_1 = 15°c

The final temperature = T_2  = 100°c

The latent heat of vaporization of water = 2260.0 J/g

Let The amount of heat required to raise temperature = Q Joule

Now, From method

Heat = mass × latent heat × change in temperature

Or, Q = m × s × ΔT

or, Q =  m × s × ( T_2 - T_1 )

So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )

Or, Q =  50 g × 2260.0 J/g × 85°c

∴   Q = 9,605,000  joule

Or, Q =  9,605 × 10³ joule

Or, Q = 9605 kilo joule

Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer

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12x+7<−11 AND5x−8≥4012
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12x

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and

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Orly uses 2 cups of raisins for every 12 cups of trail mix she makes. How many cups of trail mix will she make if she uses 8 cup
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2 years ago
The temperature at a point (x, y) on a flat metal plate is given by T(x, y) = 88/(2 + x2 + y2), where T is measured in °C and x,
-Dominant- [34]

Answer:

D_uT(3,1)=-\frac{44}{9}*\frac{1}{\sqrt{2} } \approx-3.46

Step-by-step explanation:

To find the rate of change of temperature with respect to distance at the point (3, 1) in the x-direction and the y-direction we need to find the Directional Derivative of T(x,y). The definition of the directional derivative is given by:

D_uT(x,y)=T_x(x,y)i+T_y(x,y)j

Where i and j are the rectangular components of a unit vector. In this case, the problem don't give us additional information, so let's asume:

i=\frac{1}{\sqrt{2} }

j=\frac{1}{\sqrt{2} }

So, we need to find the partial derivative with respect to x and y:

In order to do the things easier let's make the next substitution:

u=2+x^2+y^2

and express T(x,y) as:

T(x,y)=88*u^{-1}

The partial derivative with respect to x is:

Using the chain rule:

\frac{\partial u}{\partial x}=2x

Hence:

T_x(x,y)=88*(u^{-2})*\frac{\partial u}{\partial x}

Symplying the expression and replacing the value of u:

T_x(x,y)=\frac{-176x}{(2+x^2+y^2)^2}

The partial derivative with respect to y is:

Using the chain rule:

\frac{\partial u}{\partial y}=2y

Hence:

T_y(x,y)=88*(u^{-2})*\frac{\partial u}{\partial y}

Symplying the expression and replacing the value of u:

T_y(x,y)=\frac{-176y}{(2+x^2+y^2)^2}

Therefore:

D_uT(x,y)=(\frac{1}{\sqrt{2} } )*(\frac{-176x}{(2+x^2+y^2)^2} -\frac{176y}{(2+x^2+y^2)^2})

Evaluating the point (3,1)

D_uT(3,1)=(\frac{1}{\sqrt{2} } )*(\frac{-176(3)-176(1)}{(2+3^2+1^2)^2})=(\frac{1}{\sqrt{2} })* (-\frac{704}{144})=(\frac{1}{\sqrt{2} }) ( - \frac{44}{9})\approx -3.46

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