A bathtub has 62.7 kg of water at 31.0 degrees Celsius. How much water at 76.0 degrees Celsius must be added to raise the temperature at 40.3 degrees Celsius.
Water C=4186 J
Calorimetry?
2018-01-12 7:07 am
回答 (1)
2018-01-12 4:12 pm
Heat change = m c ΔT
Heat absorbed by the water at 31.0°C = Heat released by the water at 76°C
(62.7 kg) × c × [(40.3 - 31.0)°C] = m × c × [(76.0 - 40.3)°C]
(62.7 kg) × (9.3°C) = m × (35.7°C)
Mass of water at 76°C, m = 62.7 × 9.3 / 35.7 kg = 16.3 kg
Heat absorbed by the water at 31.0°C = Heat released by the water at 76°C
(62.7 kg) × c × [(40.3 - 31.0)°C] = m × c × [(76.0 - 40.3)°C]
(62.7 kg) × (9.3°C) = m × (35.7°C)
Mass of water at 76°C, m = 62.7 × 9.3 / 35.7 kg = 16.3 kg
收錄日期: 2021-04-18 18:01:35
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