Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Get immediate and reliable answers to your questions from a community of experienced experts on our platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
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°
We hope this was helpful. Please come back whenever you need more information or answers to your queries. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
