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A beaker with a heating coil contains 40.0 cm³ of liquid A. With heater power 4.80 W, the temperature rises from 15.0 deg C to 35.0 deg C in 400 s. The experiment is repeated with 20.0 cm³ of liquid A mixed with 20.0 cm³ of liquid B; with heater power 4.90 W, the temperature again rises from 15.0 deg C to 35.0 deg C in 400 s. Find the specific heat capacity c of liquid B in J kg⁻¹ K⁻¹. (Density of A = 1.60 * 10³ kg m⁻³, specific heat of A = 8.60 * 10² J kg⁻¹ K⁻¹, density of B = 2.00 * 10³ kg m⁻³)
- 8.0 * 10²
- 1.0 * 10³
- 1.2 * 10³
- 1.4 * 10³
Correct answer: 8.0 * 10²
Solution
Q1 = 4.80*400 = 1920 J; m_A1 = 0.064 kg; m_A1*c_A*dT = 0.064*860*20 = 1100.8 J; H*dT = 819.2 J; H = 40.96 J/K. Q2 = 4.90*400 = 1960 J; m_A2 = 0.032 kg; m_B = 0.040 kg. Equation: (m_A2*c_A + m_B*c_B + H)*20 = 1960 => 0.040*c_B = 29.52 => c_B = 738 J kg⁻¹ K⁻¹. Among the given options 8.0*10² is the closest.
Related JEE Advanced Physics questions
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