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3 years ago

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The addition of 9.0×105 J is required to convert a block of ice at -10 ∘C to water at 11 ∘C. i need help this is due in less tha

n 15 minutes

1 answer:

Law Incorporation [45]3 years ago

8 0

The mass of the block of ice is 2.24 kg

Explanation:

The amount of heat needed for the whole process consists of three different amounts of heat:

$Q_1$ : the amount of heat needed to raise the temperature of the block of ice from $-10^{\circ}C$ to $0^{\circ}C$

$Q_2$ : the amount of heat needed to melt the block of ice at melting point

$Q_3$ : the amount of heat needed to raise the temperature of the water from $0^{\circ}C$ to $11^{\circ}C$

The total amount of heat needed can be written as

$Q=Q_1+Q_2+Q_3=mC_i\Delta T_1 + m\lambda_f + mC_w\Delta T_3$

where we have:

$Q=9.0 \cdot 10^5 J$ (total amount of heat required)

m is the mass of the block of ice

$C_i = 2108 J/kg^{\circ}C$ is the specific heat of ice

$\lambda_f=3.34\cdot 10^5 J/kg$ is the latent heat of fusion of ice

$C_w=4186 J/kg^{\circ}C$ is the specific heat capacity of water

$\Delta T_1 = 0-(-10)=10^{\circ}C$ is the change in temperature in the 1st process

$\Delta T_3 = 11-0=11^{\circ}C$ is the change in temperature in the 3rd process

Solving the equation for m, we find the mass of the block of ice:

$m=\frac{Q}{C_i\Delta T_1 + \lambda_f+C_w\Delta T_3}=\frac{9.0\cdot 10^5}{(2108)(10)+3.34\cdot 10^5+(4186)(11)}=2.24 kg$

Learn more about specific heat capacity:

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Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

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$h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}$

$h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}$

Our interest value would be given using interpolation like this:

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2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that $\dot{m}=\frac{\dot{V}}{\upsilon}$ , but since $\dot{V}=Av$ we have this:

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$v=\frac{\upsilon \dot{m}}{A}$ (*)

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$v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}$

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