A 47.70 g sample of a substance is initially at 20.7 °C. After absorbing 1271 J of heat, the temperature of the substance is 119.1 °C?
2021-03-24 12:17 pm
A 47.70 g sample of a substance is initially at 20.7 °C. After absorbing 1271 J of heat, the temperature of the substance is 119.1 °C. What is the specific heat ( 𝑐 ) of the substance?
回答 (3)
2021-03-24 2:26 pm
✔ 最佳答案
q = m c ΔT1271 J = (47.70 g) × c × [(119.1 - 20.7) °C]
c = 1271 / [47.70 × (119.1 - 20.7)] J/g°C
Specific heat, c = 0.271 J/g°C
2021-03-24 12:58 pm
Use the same equation used in some of the other questions you've asked:
q = m c (T2-T1)
Plug in what you have and solve for c.
q = m c (T2-T1)
Plug in what you have and solve for c.
2021-03-24 12:41 pm
Q = m*c*delta-T.
In your problem, Q = 1271 J, m = 47.70 g, and delta-T = 98.6 K.
So c = [1271/(47.70*98.6)] J/gK = around 0.28 J/gK but use a calculator.
In your problem, Q = 1271 J, m = 47.70 g, and delta-T = 98.6 K.
So c = [1271/(47.70*98.6)] J/gK = around 0.28 J/gK but use a calculator.
收錄日期: 2021-04-23 23:07:11
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