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Use the following information to answer the next question:

A thermite reaction is a highly exothermic process that is used in welding massive steel objects such as ship propellers and train rails. The reaction can be represented by the following equation:

[tex]\[ 2 \text{Al}(s) + \text{Fe}_2\text{O}_3(s) \rightarrow \text{Al}_2\text{O}_3(s) + 2 \text{Fe}(l) \quad \Delta H = -851.5 \, \text{kJ} \][/tex]

If the heat produced by the reaction of [tex]\(1.00 \, \text{mol}\)[/tex] of iron(III) oxide were absorbed by [tex]\(7.40 \, \text{kg}\)[/tex] of [tex]\(\text{H}_2\text{O}(l)\)[/tex] at room temperature, what would be the resulting temperature change of the water?

The amount of heat released by a substance when the temperature of [tex]\(250.0 \, \text{g}\)[/tex] of water increases from [tex]\(20.0^\circ \text{C}\)[/tex] to [tex]\(35.0^\circ \text{C}\)[/tex] is _____.

Sagot :

GinnyAnswer
Sure, let's solve the given problem step by step:

Question: What is the amount of heat released when the temperature of 250.0 g of water increases from 20.0°C to 35.0°C?

### Step-by-Step Solution:

1. Determine the mass of the water [tex]\( (m) \)[/tex]:
The mass of the water given is 250.0 grams.

2. Identify the initial and final temperatures [tex]\( (T_i \text{ and } T_f) \)[/tex]:
- Initial temperature [tex]\( T_i \)[/tex] = 20.0°C
- Final temperature [tex]\( T_f \)[/tex] = 35.0°C

3. Calculate the temperature change [tex]\( (\Delta T) \)[/tex]:
[tex]\[ \Delta T = T_f - T_i \][/tex]
Substitute:
[tex]\[ \Delta T = 35.0°C - 20.0°C = 15.0°C \][/tex]

4. Find the specific heat capacity of water [tex]\( (c) \)[/tex]:
The specific heat capacity of water is 4.18 J/g°C.

5. Use the formula for the amount of heat [tex]\( (q) \)[/tex]:
The formula to calculate the amount of heat absorbed or released is:
[tex]\[ q = m \cdot c \cdot \Delta T \][/tex]

6. Substitute the values into the formula:
- [tex]\( m = 250.0 \)[/tex] grams
- [tex]\( c = 4.18 \)[/tex] J/g°C
- [tex]\( \Delta T = 15.0 \)[/tex]°C

[tex]\[ q = 250.0 \, \text{g} \times 4.18 \, \text{J/g°C} \times 15.0 \, \text{°C} \][/tex]

7. Calculate the result:
[tex]\[ q = 250.0 \times 4.18 \times 15.0 \][/tex]
[tex]\[ q = 15675.0 \, \text{J} \][/tex]

### Final Answer:
The amount of heat released when the temperature of 250.0 grams of water increases from 20.0°C to 35.0°C is 15,675.0 Joules.
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