Values and Equations:
Specific heat of water (gas) = 2.01 J/g°C
Specific heat of water (liquid) = 4.184 J/g°C
Specific heat of water (solid) = 2.03 J/g°C
Hf water = 333.5 J/g
Hv water = 2256.8 J/g
q=mc∆T
q=mHf
q=mHv

Calculate the amount of energy in joules that must be added to 245.0 g of ice at -45.0 ºC to turn it all into liquid water at 0.00 ºC.  Hint, two separate calculations must be done and then added together.


To calculate the amount of energy required to turn 245.0 g of ice at -45.0 ºC into liquid water at 0.00 ºC, we need to do the following two steps:

### Step 1: Heat the ice from -45.0 ºC to 0.00 ºC
This step involves increasing the temperature of the ice without changing its phase. We use the equation:

\[
q_1 = mc\Delta T
\]

Where:
- \(m = 245.0 \, \text{g}\) (mass of the ice),
- \(c = 2.03 \, \text{J/g°C}\) (specific heat of ice),
- \(\Delta T = 0.00 - (-45.0) = 45.0 \, \text{°C}\) (temperature change).

\[
q_1 = (245.0 \, \text{g}) \times (2.03 \, \text{J/g°C}) \times (45.0 \, \text{°C})
\]

\[
q_1 = 22417.25 \, \text{J}
\]

### Step 2: Melt the ice at 0.00 ºC into liquid water
This step involves changing the phase from ice to liquid water at the same temperature (0.00 ºC). For this, we use the heat of fusion equation:

\[
q_2 = mH_f
\]

Where:
- \(m = 245.0 \, \text{g}\),
- \(H_f = 333.5 \, \text{J/g}\) (heat of fusion of ice).

\[
q_2 = (245.0 \, \text{g}) \times (333.5 \, \text{J/g})
\]

\[
q_2 = 81707.5 \, \text{J}
\]

### Total Energy Required
Now, we sum the energy from both steps:

\[
q_{\text{total}} = q_1 + q_2 = 22417.25 \, \text{J} + 81707.5 \, \text{J}
\]

\[
q_{\text{total}} = 104124.75 \, \text{J}
\]

So, the total amount of energy required is approximately **104,125 J**.

Would you like further details or clarifications?

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Here are 5 related questions to expand on this:
1. What is the amount of energy required to heat the same mass of liquid water from 0.00 ºC to 100.0 ºC?
2. How much energy is needed to convert 245.0 g of liquid water at 100.0 ºC to steam?
3. What is the difference in energy required to melt 100 g of ice compared to boiling 100 g of water?
4. If the ice was at -10.0 ºC instead of -45.0 ºC, how much energy would be needed to melt it completely?
5. How does the specific heat of ice compare to the specific heat of steam, and what implications does that have for energy absorption?

**Tip**: Always keep track of phase changes in thermodynamic problems, as different equations apply to heating/cooling and phase transitions.

Calculate the amount of energy in joules that must be added to 245.0 g of ice at -45.0 ºC to turn it all into liquid water at 0.00 ºC. Hint, two separate calculations must be done and then added together.

Discover the steps to calculate the total energy required to convert 245.0 g of ice at -45.0 ºC into liquid water at 0.00 ºC. This problem covers key thermodynamic principles, including energy calculations using specific heat and heat of fusion. Suitable for students in Grades 10-12.

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