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.
