Calculate the heat absorbed (in joules) by 15.1 g of water to raise its temperature from 22 degrees Celsius to 51 degrees Celsius (at constant pressure). What information do we need to calculate the heat absorbed? How much heat will be absorbed to raise 15.1 g of water by one degree Celsius? However, we want to raise the temperature of our sample of water by more than 1 degree Celsius. What is the temperature change that occurs? Now calculate the heat absorbed by 15.1 g of water to raise its temperature from 22 degrees Celsius to 51 degrees Celsius (at constant pressure).
To calculate the heat absorbed by 15.1 g of water when its temperature increases, we use the formula: q=mcΔTq = mc\Delta Tq=mcΔT
Where:
- qqq is the heat absorbed (in joules),
- mmm is the mass of the water (in grams),
- ccc is the specific heat capacity of water (4.18 J/g°C),
- ΔT\Delta TΔT is the change in temperature (in °C).
Information needed to calculate heat absorbed:
- Mass of the water (m): 15.1 g,
- Specific heat capacity of water (c): 4.18 J/g°C,
- Temperature change (ΔT\Delta TΔT): The difference between the final and initial temperatures, which is 51−22=2951 – 22 = 2951−22=29 °C.
Heat absorbed to raise 15.1 g of water by 1°C:
Using the formula q=mcΔTq = mc\Delta Tq=mcΔT and setting ΔT=1\Delta T = 1ΔT=1, we can calculate the heat absorbed per 1°C rise: q=15.1 g×4.18 J/g°C×1 °C=63.2 Jq = 15.1 \, \text{g} \times 4.18 \, \text{J/g°C} \times 1 \, \text{°C} = 63.2 \, \text{J}q=15.1g×4.18J/g°C×1°C=63.2J
Thus, 63.2 joules of heat will be absorbed to raise the temperature of 15.1 g of water by 1°C.
Temperature change that occurs:
We are raising the temperature of the water from 22°C to 51°C, which gives a temperature change: ΔT=51−22=29 °C\Delta T = 51 – 22 = 29 \, \text{°C}ΔT=51−22=29°C
Heat absorbed to raise the temperature from 22°C to 51°C:
Now, we can calculate the total heat absorbed to raise the temperature by 29°C: q=15.1 g×4.18 J/g°C×29 °C=1746.3 Jq = 15.1 \, \text{g} \times 4.18 \, \text{J/g°C} \times 29 \, \text{°C} = 1746.3 \, \text{J}q=15.1g×4.18J/g°C×29°C=1746.3J
Thus, the heat absorbed by 15.1 g of water to raise its temperature from 22°C to 51°C is 1746.3 joules.
Final Answer:
- The heat absorbed to raise 15.1 g of water by 1°C is 63.2 joules.
- The total heat absorbed to raise the temperature from 22°C to 51°C is 1746.3 joules.
This calculation assumes no phase changes and that the water remains in liquid form throughout the temperature increase.
