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How many kilojoules of energy are needed to convert 280 grams of water at -20 C° to water at 75 C°? (3 steps-water to freezing, phase change, water to 75 C°)

Sagot :

INFORMATION:

We need to find how many kilojoules of energy are needed to convert 280 grams of water at -20 C° to water at 75 C°

STEP BY STEP EXPLANATION:

Since the temperature of water in all process goes from -20°C to 75°C, we would have a change in the state when temperature is equal to 0°C.

So, we must divide the problem in three parts:

1. Temperature from -20°C to 0°C:

In this part, we need to use the next formula

[tex]Q=m\times C_s\timesΔT[/tex]

Where, m represent the mass, Cs is the specific heat capacity and ΔT is the temperature change.

In this case,

m = 280 grams = 0.28 Kg

Cs = 2100 J/Kg°C

ΔT = 0°C - (-20 °C) = 20°C

Now, replacing in the formula

[tex]\begin{gathered} Q=0.28Kg\times2100\frac{J}{Kg\cdot\degree C}\times20\degree C \\ Q=11760J \end{gathered}[/tex]

2. Change in the state from ice to water:

In this part, we need to use the next formula

[tex]Q=m\times L[/tex]

Where, m represents the mass and L is the specific latent heat.

In this case,

m = 280 grams = 0.28 Kg

L = 334 J/Kg

Now, replacing in the formula

[tex]\begin{gathered} Q=0.28Kg\times334\frac{J}{Kg} \\ Q=93.52J \end{gathered}[/tex]

3. Temperature from 0°C to 75°C:

In this part, we need to use the formula from the first part.

In this case,

m = 280 grams = 0.28 Kg

Cs = 4186 J/Kg°C

ΔT = 75°C - 0 °C = 75°C

Now, replacing in the formula

[tex]\begin{gathered} Q=0.28Kg\times4186\frac{J}{Kg\operatorname{\degree}C}\times75\operatorname{\degree}C \\ Q=87906J \end{gathered}[/tex]

Finally, we must add up the three parts and then convert it to kilojoules.

[tex]\begin{gathered} 11760J+93.52J+87906J=99759.52J \\ \text{ To kilojoules,} \\ =99.7595KJ \end{gathered}[/tex]

ANSWER:

99.7595 kilojoules of energy are needed to convert 280 grams of water at -20 C° to water at 75 C°

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