This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Step 1: Define specific heat capacity. Specific heat capacity is the quantity of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius or one Kelvin. It indicates how much heat a substance can absorb before its temperature changes significantly. Step 2: Calculate the specific heat capacity of paraffin. Given: Heat supplied, Q = 21600 \, J Mass of paraffin, m = 2.0 \, kg Temperature rise, T = 4.9 \, °C The formula for heat transfer is Q = mc T. Rearrange to solve for specific heat capacity c: c = (Q)/(m T) Substitute the given values: c = 21600 \, J(2.0 \, kg)(4.9 \, °C) c = 21600 \, J9.8 \, kg°C c = 2204.08 \, Jkg^-1°C^-1 Rounding to three significant figures: c = 2200 \, Jkg^-1°C^-1 Step 3: State with reason the flask in which the water is likely to have a higher temperature eight hours later. Flask A, which is partially filled, will likely have a higher temperature eight hours later. The air trapped above the water in flask A acts as an insulating layer, reducing heat loss by convection and conduction from the water surface. In flask B, which is completely filled, there is less or no air space, allowing for more direct heat transfer to the flask walls and surroundings. Step 4: Determine the final temperature of the mixture. Given: Mass of ice, m_ice = 40 \, g = 0.040 \, kg Initial temperature of ice, T_ice = 0 \, °C Mass of water, m_water = 400 \, g = 0.400 \, kg Initial temperature of water, T_water = 20 \, °C Specific heat capacity of water, c_water = 4200 \, Jkg^-1K^-1 Specific latent heat of fusion of ice, L_f = 340000 \, Jkg^-1 Let the final temperature be T_f. Heat gained by ice to melt (Q_1): Q_1 = m_ice L_f Q_1 = (0.040 \, kg)(340000 \, Jkg^-1) Q_1 = 13600 \, J Heat gained by melted ice (now water) to reach T_f (Q_2): Q_2 = m_ice c_water (T_f - 0 \, °C) Q_2 = (0.040 \, kg)(4200 \, Jkg^-1°C^-1) T_f Q_2 = 168 T_f \, J Heat lost by initial water to reach T_f (Q_3): Q_3 = m_water c_water (T_water - T_f) Q_3 = (0.400 \, kg)(4200 \, Jkg^-1°C^-1) (20 \, °C - T_f) Q_3 = 1680 (20 - T_f) \, J Q_3 = 33600 - 1680 T_f \, J According to the principle of calorimetry, heat gained equals heat lost: Q_1 + Q_2 = Q_3 13600 + 168 T_f = 33600 - 1680 T_f Combine terms with T_f: 168 T_f + 1680 T_f = 33600 - 13600 1848 T_f = 20000 T_f = (20000)/(1848) T_f ≈ 10.82 \, °C Rounding to three significant figures: T_f = 10.8 \, °C Step 5: Determine the linear velocity. Given: Frequency, f = 3 \, revolutions per second Radius, r = 1.5 \, m First, calculate the angular velocity : = 2 f = 2 (3 \, s^-1) = 6 \, rad/s Next, calculate the linear velocity v: v = r v = (6 \, rad/s)(1.5 \, m) v = 9 \, m/s Using ≈ 3.14159: v ≈ 9 × 3.14159 \, m/s v ≈ 28.274 \, m/s Rounding to three significant figures: v = 28.3 \, m/s Send me the next one 📸