Q3 V3If the temperature of 100 mL of solution increased by 3 Kelvin (K) during the reaction and the specific heat capacity of the solution is 4.18 J/g°C, how much heat energy was absorbed?

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BlackTom AI · Accepted Answer

First, identify the formula for heat transfer in a solution: Q = m × c × ΔT, where m is mass, c is specific heat capacity, and ΔT is the change in temperature.
Next, determine the mass. A volume of 100 mL of solution is approximately 100 g if the density is near 1 g/mL, which is a common assumption for dilute aqueous solutions used in these problems. So, m ≈ 100 g.
Then, use the given specific heat capacity: c = 4.18 J/g·°C.
The temperature change ΔT is 3 K. Since the size of the degree is the same in Kelvin and Celsius, ΔT = 3 °C (they are equivalent for a size change).
Now plug these values into the formula: Q = (100 g) × (4.18 J/g·°C) × (3 °C).
Multiplying 4.18 × 3 gives 12.54, and then 12.54 × 100 g yields 1254 J.
Evaluate each answer option:
- a. 125.4 J: This is off by a factor of 10, likely due to misplacing the decimal during the multiplication or miscounting the mass (10 g vs. 100 g). It does not match the calculated result.
- b. 12540 J: This would correspond to a tenfold increase in either mass or ΔT, which is not supported by the given values. It overestimates the energy by a factor of 10.
- c. 1254 J: This matches the calculation exactly when m = 100 g, c = 4.18 J/g·°C, and ΔT = 3 °C, so the math aligns with the provided data.
- d. Cannot calculate as the temperature change is units of Kelvin (K): Here, a Kelvin change has the same magnitude as a Celsius change for ΔT, so using 3 K is equivalent to 3 °C in this context. The units do not prevent calculation.
In summary, the correct calculation yields 1254 J, which corresponds to the option that lists 1254 J. The other choices either misapply the mass, misplace the decimal, or rely on a misleading interpretation of units, leading to incorrect results.

MUF0042 Chemistry Unit 2 - Semester 2, 2025

Q3 V3If the temperature of 100 mL of solution increased by 3 Kelvin (K) during the reaction and the specific heat capacity of the solution is 4.18 J/g°C, how much heat energy was absorbed?

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