Вопрос
b) (1). A circular copper sing at 30.0^circ C has a hole with an area of 21.3 cm^2 . What minimum temperature must it have so that it can be slippe onto a steel metal rod having a cross -sectional area of 22.00cm^2 (a copper is 17times 10^-6 and a steel is 11times 10^-6) (5marks) (ii). Suppose the ring and the rod are heated simultaneously. Wh minimum change in temperature of both will allow the ring to be onto the end of the rod. (5marks)
Решения
4.5
(162 Голоса)
Чулпан
ветеран · Репетитор 11 лет
Ответ
To solve this problem, we need to use the concept of thermal expansion and the principle of thermal equilibrium.(i) To find the minimum temperature at which the copper ring can be slipped onto the steel metal rod, we need to ensure that the expansion of the copper ring is equal to the expansion of the steel rod.The formula for thermal expansion is:ΔL = α * L * ΔTwhere ΔL is the change in length, α is the coefficient of linear expansion, L is the original length, and ΔT is the change in temperature.For the copper ring, the expansion is given by:ΔL_copper = α_copper * L_copper * ΔTFor the steel rod, the expansion is given by:ΔL_steel = α_steel * L_steel * ΔTTo ensure that the copper ring can be slipped onto the steel rod, the expansion of the copper ring must be equal to the expansion of the steel rod. Therefore, we have:ΔL_copper = ΔL_steelSubstituting the values, we get:α_copper * L_copper * ΔT = α_steel * L_steel * ΔTSolving for ΔT, we get:ΔT = (α_steel * L_steel) / (α_copper * L_copper)Substituting the given values, we get:ΔT = (11 × 10^-6 * L_steel) / (17 × 10^-6 * L_copper)(ii) To find the minimum change in temperature that will allow the ring to be slipped onto the end of the rod, we need to ensure that the expansion of the copper ring is equal to the expansion of the steel rod.Using the same formula as before, we have:ΔL_copper = α_copper * L_copper * ΔTΔL_steel = α_steel * L_steel * ΔTTo ensure that the copper ring can be slipped onto the end of the steel rod, the expansion of the copper ring must be equal to the expansion of the steel rod. Therefore, we have:ΔL_copper = ΔL_steelSubstituting the given values, we get:α_copper * L_copper * ΔT = α_steel * L_steel * ΔTSolving for ΔT, we get:ΔT = (α_steel * L_steel) / (α_copper * L_copper)Substituting the given values, we get:ΔT = (11 × 10^-6 * L_steel) / (17 × 10^-6 * L_copper)Therefore, the minimum temperature that the copper ring must have so that it can be slipped onto the steel metal rod is given by the above expression. Similarly, the minimum change in temperature that will allow the ring to be slipped onto the end of the rod is also given by the same expression.