Calculate the heat absorbed in Joules when 32 grams of aluminum are heated from 28C to 85C using q= m x C x temp change
The Correct Answer and Explanation is:
To calculate the heat absorbed by aluminum, we will use the formula: q=m×C×ΔTq = m \times C \times \Delta Tq=m×C×ΔT
Where:
- qqq is the heat absorbed (in Joules)
- mmm is the mass of the substance (in grams)
- CCC is the specific heat capacity of aluminum (in Joules per gram per degree Celsius)
- ΔT\Delta TΔT is the change in temperature (in degrees Celsius)
Given:
- Mass of aluminum, m=32 gm = 32 \, \text{g}m=32g
- Specific heat capacity of aluminum, C=0.897 J/g°CC = 0.897 \, \text{J/g°C}C=0.897J/g°C (This value is commonly used for aluminum)
- Temperature change, ΔT=Tfinal−Tinitial=85°C−28°C=57°C\Delta T = T_{\text{final}} – T_{\text{initial}} = 85°C – 28°C = 57°CΔT=Tfinal−Tinitial=85°C−28°C=57°C
Calculation:
Now, substituting the values into the formula: q=32 g×0.897 J/g°C×57°Cq = 32 \, \text{g} \times 0.897 \, \text{J/g°C} \times 57°Cq=32g×0.897J/g°C×57°C q=32×0.897×57q = 32 \times 0.897 \times 57q=32×0.897×57 q=32×51.069q = 32 \times 51.069q=32×51.069 q=1634.21 Jq = 1634.21 \, \text{J}q=1634.21J
Thus, the heat absorbed by the aluminum is 1634.21 Joules.
Explanation:
In this problem, we are using the specific heat capacity to determine how much heat is required to raise the temperature of the aluminum by 57°C. The specific heat capacity tells us how much heat is needed to raise the temperature of one gram of a substance by one degree Celsius. By multiplying the mass of the aluminum by its specific heat and the temperature change, we can calculate the total heat absorbed.
This calculation assumes that the specific heat capacity remains constant over the temperature range, which is generally a valid assumption for small temperature changes. The result tells us how much energy in the form of heat is required to heat up the given mass of aluminum from 28°C to 85°C.
