The specific heat of ice is 2.1 kJ/kg.°C. When 20 kJ of heat is removed from 1.2 kg of ice originally at -15°C, its new temperature is

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BlackTom AI · Accepted Answer

We start with the given data and the process described.
- Mass m = 1.2 kg
- Specific heat of ice c = 2.1 kJ/(kg·°C)
- Initial temperature T_i = -15°C
- Heat removed ΔQ = 20 kJ (since heat is being taken away, the temperature will drop)

Step 1: Determine the temperature change if all the removed heat goes into lowering the temperature of the ice without phase change.
- For heating or cooling without phase change, the relation ΔQ = m · c · ΔT applies.
- Here, ΔQ = 20 kJ, m = 1.2 kg, c = 2.1 kJ/(kg·°C).
- Solve for ΔT: ΔT = ΔQ / (m·c) = 20 / (1.2 × 2.1) = 20 / 2.52 ≈ 7.94°C.

Step 2: Apply the temperature change to the initial temperature.
- Since heat is removed, the temperature decreases: T_f = T_i − ΔT = −15°C − 7.94°C ≈ −22.94°C ≈ −23°C.

Analysis of the answer options:
- 35°C: This would require adding heat and a large temperature rise, which contradicts the given condition of removing heat. It is incorrect.
- –23°C: This matches the calculated final temperature after removing 20 kJ of heat from the ice, considering no phase change occurs during the process. This is correct.
- –26°C: This would correspond to a larger temperature drop (ΔT ≈ 11°C), which would require more than 20 kJ of removal given the same mass and specific heat. It is incorrect.
- –18°C: This would imply a much smaller temperature drop (ΔT ≈ 3°C), inconsistent with the actual ΔT calculated from the given heat removal. It is incorrect.

In summary, the only physically consistent result with 20 kJ removed from 1.2 kg of ice at −15°C, using c = 2.1 kJ/(kg·°C) and without phase change, is a final temperature of about −23°C.

Win25_PHYS010_10428 EC Quiz Ch.7-8

The specific heat of ice is 2.1 kJ/kg.°C. When 20 kJ of heat is removed from 1.2 kg of ice originally at -15°C, its new temperature is

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