part a a calorimeter consists of an aluminum cup inside of an insulated container. the cup is weighed on a top-loading balance and is found to have a mass of 31.91 g. a reaction is conducted in the calorimeter, raising the temperature from 21.2 c to 26.1 c what is the change in heat q for the aluminum cup in units of j? aluminum has a specific heat of 0.903 j 1 ? 1 write your answer to the correct number of significant figures.
What is the change in heat ( q ) for the aluminum cup in units of joules (J) when the temperature is raised from 21.2°C to 26.1°C, given a mass of 31.91 g and a specific heat of 0.903 J/g°C?
Answer:
To calculate the change in heat (( q )) for the aluminum cup, we can use the formula for heat transfer:
q = mc\Delta T
where:
- ( m ) is the mass of the aluminum cup
- ( c ) is the specific heat capacity of aluminum
- ( \Delta T ) is the change in temperature
Let’s break the calculation into steps:
-
Determine the mass ( m ):
- The mass of the aluminum cup is given as m = 31.91 \, \text{g}
-
Identify the specific heat capacity ( c ):
- For aluminum, the specific heat capacity is c = 0.903 \, \text{J/g°C}
-
Calculate the change in temperature ( \Delta T ):
-
The temperature changes from 21.2 \, ^\circ\text{C} to 26.1 \, ^\circ\text{C}
\Delta T = T_{\text{final}} - T_{\text{initial}} = 26.1 \, ^\circ\text{C} - 21.2 \, ^\circ\text{C} = 4.9 \, ^\circ\text{C}
-
-
Calculate the change in heat ( q ):
q = mc\Delta T = (31.91 \, \text{g})(0.903 \, \text{J/g°C})(4.9 \, ^\circ\text{C}) -
Perform the calculation:
q = 31.91 \times 0.903 \times 4.9q \approx 141.2 \, \text{J}
Since the given data has:
- 31.91 (four significant figures)
- 0.903 (three significant figures)
- 4.9 (two significant figures)
The final answer must be rounded to the least number of significant figures, which is 2 significant figures in this case.
Final Answer:
The change in heat ( q ) for the aluminum cup is approximately \boxed{140 \, \text{J}}