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Physics · Heat, temperature, thermal expansion, specific heat capacity, calorimetry, change of state, latent heat

A electric heater is used to heat a container filled with of water. It is found

A \( 10 W \) electric heater is used to heat a container filled with \( 0.5 \mathrm{kg} \) of water. It is found that the temperature of water and the container rises by \( 3 K \) in 15 min. The container is then emptied, dried and filled with \( 2 k g \) of oil. The same heater now raises the temperature of container-oil system by \( 2 K \) in 20 min. Assuming that there is no heat loss in the process and the specific heat of water is \( 4200 \mathrm{Jkg}^{-1} \mathrm{K}^{-1} \), the specific heat of oil in the same unit is equal to:

  • A. \( 1.50 \times 10^{3} \)
  • B. \( 2.55 \times 10^{3} \)
  • C. \( 3.00 \times 10^{3} \)
  • D. \( 5.10 \times 10^{3} \)

Step-by-step solution

Heater power P = 10 W. For water: mass m_w = 0.5 kg, ΔT_w = 3 K, time t1 = 15 min = 900 s. Heat supplied Q1 = P t1 = 10 × 900 = 9000 J. This heats water and container: Q1 = m_w c_w ΔT_w + C_c ΔT_w, where C_c is container heat capacity. Solving: 9000 = 0.5 × 4200 × 3 + C_c × 3 => 9000 = 6300 + 3C_c => C_c = 900 J/K. For oil: mass m_o = 2 kg, ΔT_o = 2 K, time t2 = 20 min = 1200 s. Heat supplied Q2 = 10 × 1200 = 12000 J. This heats oil and container: Q2 = m_o c_o ΔT_o + C_c ΔT_o => 12000 = 2 × c_o × 2 + 900 × 2 => 12000 = 4c_o + 1800 => 4c_o = 10200 => c_o = 2550 J/kgK = 2.55 × 10^3 J/kgK.
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