An insulated calorimeter contains 100 g of water at 10 °C. A salt is added to the solution, causing the temperature to drop by 3.0 °C. Given that water has a specific heat of 4.184 J/g°C, how much heat was transferred to or from the water?Group of answer choices1.3 kJ3.4 kJ4.2 kJ2.9 kJ
Question
An insulated calorimeter contains 100 g of water at 10 °C. A salt is added to the solution, causing the temperature to drop by 3.0 °C. Given that water has a specific heat of 4.184 J/g°C, how much heat was transferred to or from the water?Group of answer choices1.3 kJ3.4 kJ4.2 kJ2.9 kJ
Solution
The heat transferred can be calculated using the formula:
q = mcΔT
where:
- q is the heat transferred
- m is the mass of the substance
- c is the specific heat of the substance
- ΔT is the change in temperature
In this case:
- m = 100 g
- c = 4.184 J/g°C
- ΔT = -3.0 °C (the temperature dropped, so the change is negative)
Substituting these values into the formula gives:
q = (100 g)(4.184 J/g°C)(-3.0 °C) = -1255.2 J
Since 1 kJ = 1000 J, this is equivalent to -1.2552 kJ. However, since the question asks for the amount of heat transferred, we take the absolute value, which is 1.2552 kJ.
So, the closest answer is 1.3 kJ.
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