A chef heats some water in a pan on a hotplate. The temperature of the water rises by 10°C in time t. She then puts the same volume of oil in an identical pan on the same hotplate. The specific heat capacity of water is 2.5 times that of oil and water is 1.1 times denser than oil. What is the time for the temperature of the oil to rise by 10°C? Options: A 0.36t B 0.44t C 2.3t D 2.8t---**Question 15** **Question Stem:** A chef heats some water in a pan on a hotplate. The temperature of the water rises by 10°C in time t. She then puts the same volume of oil in an identical pan on the same hotplate. The specific heat capacity of water is 2.5 times that of oil and water is 1.1 times denser than oil. What is the time for the temperature of the oil to rise by 10°C? **Options:** A 0.36t B 0.44t C 2.3t D 2.8t **Other Relevant Text:** ρ = m/V © UCLES 2023 0625/21/O/N/23 183 [Handwritten notes, possibly working:] C = E / (m * Δt) E = C * m * Δt 0.44 t (crossed out) E / (0.4 * C * 0.9 * m) = t' (partially visible, inferred) 0.4C x 0.9m x t = E (partially visible, inferred) t = E / (0.4C x 0.9m) (partially visible, inferred) t = E / (0.36 Cm) (partially visible, inferred) t' = E / (C_oil * m_oil * Δt) Δt = 10°C (given) E is the same (same hotplate, same time, assuming constant power) [More handwritten notes, partially visible and potentially related to the calculation:] 0.4 x C x m x t = E C = E / (m x t) C_oil = C_water / 2.5 = C_water / 2.5 ρ_water = 1.1 * ρ_oil m_water = ρ_water * V = 1.1 * ρ_oil * V m_oil = ρ_oil * V m_water / m_oil = 1.1 Let C_water = C, C_oil = C / 2.5 = 0.4C Let m_water = m, m_oil = m / 1.1 ≈ 0.91m (calculation error in notes, 0.9 is written) Energy absorbed by water: E = C_water * m_water * ΔT = C * m * 10 Hotplate power = P = E / t = (C * m * 10) / t For oil: E_oil = P * t_oil = C_oil * m_oil * ΔT = (C/2.5) * (m/1.1) * 10 (C * m * 10) / t * t_oil = (C/2.5) * (m/1.1) * 10 t_oil = t * (C/2.5) * (m/1.1) * 10 / (C * m * 10) t_oil = t * (1 / 2.5) * (1 / 1.1) t_oil = t / (2.5 * 1.1) t_oil = t / 2.75 t_oil ≈ 0.3636 t [Handwritten calculation matches approximately the above derivation, using 0.4C and 0.9m (incorrect mass ratio)] C_water * m_water * 10 = P * t C * m * 10 = P * t P = 10Cm/t C_oil * m_oil * 10 = P * t_oil 0.4C * (m/1.1) * 10 = P * t_oil 0.4C * (m/1.1) * 10 = (10Cm/t) * t_oil 0.4 / 1.1 = t_oil / t t_oil = t * (0.4 / 1.1) = t * (4/11) ≈ 0.3636t [The handwritten notes seem to use m_oil = 0.9m instead of m/1.1 ≈ 0.909m] 0.4C * 0.9m * 10 = P * t_oil 0.4 * 0.9 * Cm * 10 = (10Cm/t) * t_oil 0.36 * 10 = 10/t * t_oil 3.6 = 10/t * t_oil t_oil = 3.6 * t / 10 = 0.36t [The crossed out 0.44t might be from a different calculation] [The final handwritten calculation segment seems to be based on the (incorrect) assumption m_oil = 0.9m and C_oil = 0.4C, leading to 0.36t.]