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A 50 g sample of an unknown metal is heated to 90.0C. It is placed in a perfectly insulated container along with 100 g of water at an initial temperature of 20C. After a short time, the temperature of both the metal and water become equal at 25C. The specific heat of water is 4.18 J/gC in this temperature range. What is the specific heat capacity of the metal?
Record your answer with two significant figures.
J/gC
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Answered by
GPT-5 mini
AI
Heat lost by metal = heat gained by water.
m_m c_m (T_i,m - T_f) = m_w c_w (T_f - T_i,w)
50 g · c_m · (90.0 − 25.0) = 100 g · 4.18 J/g°C · (25.0 − 20.0)
50 · c_m · 65 = 100 · 4.18 · 5 = 2090 J
c_m = 2090 / (50·65) = 0.643 J/g°C ≈ 0.64 J/g°C (two significant figures).
m_m c_m (T_i,m - T_f) = m_w c_w (T_f - T_i,w)
50 g · c_m · (90.0 − 25.0) = 100 g · 4.18 J/g°C · (25.0 − 20.0)
50 · c_m · 65 = 100 · 4.18 · 5 = 2090 J
c_m = 2090 / (50·65) = 0.643 J/g°C ≈ 0.64 J/g°C (two significant figures).
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