M1, г -250 t1, °С -80 m2, г -165 t2, °С -(-5) t3, °С -15 св, -4200 сл 2100 Дж/(кг·оС). Дж/(кг·оС). эту теплоту c) На плавление льда затрачивается теплота Q3, которую можно рассчитать по формуле d) талая вода (бывший лед) нагревается до конечной температуры смеси, для этого он должен получить теплоту Q4. Рассчитала ее по формуле уравнение теплового баланса имеет вид: Q1 = Q2 + Q3 + Q4, откуда Вычислите и ответ запишите в кДж/кг 8) Сравнила полученное значение с табличным: λ=335 кДж/кг
Given: m1 = -250 g (mass of first substance) t1 = -80 °C (temperature of first substance) m2 = -165 g (mass of second substance) t2 = -(-5) °C (temperature of second substance) t3 = -15 °C (temperature of final mixture) св = -4200 J/(kg·°C) (specific heat capacity of first substance) сл = 2100 J/(kg·°C) (specific heat capacity of second substance)
a) To find Q1, the heat transferred to the first substance, we use the formula: Q1 = m1 * св * (t3 - t1) = -250 g * -4200 J/(kg·°C) * (-15 °C - (-80 °C)) = -250 g * -4200 J/(kg·°C) * 65 °C = 68250000 J
Q1 = 68250000 J = 68.25 kJ
b) To find Q2, the heat transferred to the second substance, we use the formula: Q2 = m2 * сл * (t3 - t2) = -165 g * 2100 J/(kg·°C) * (-15 °C - (-5 °C)) = -165 g * 2100 J/(kg·°C) * (-10 °C) = 3465000 J
Q2 = 3465000 J = 3.465 kJ
c) To find Q3, the heat needed for melting ice, we can use the specific heat capacity of the first substance since it undergoes a phase change: Q3 = m1 * св * (0 °C - t2) = -250 g * -4200 J/(kg·°C) * (0 °C - (-5 °C)) = -250 g * -4200 J/(kg·°C) * 5 °C = 525000 J
Q3 = 525000 J = 0.525 kJ
d) To find Q4, the heat needed to raise the melted water to the final temperature, we can use the specific heat capacity of water: Q4 = m1 * сл * (t3 - 0 °C) = -250 g * 2100 J/(kg·°C) * (-15 °C - 0 °C) = -250 g * 2100 J/(kg·°C) * (-15 °C) = 787500 J
Q4 = 787500 J = 0.7875 kJ
The equation for the heat balance is: Q1 = Q2 + Q3 + Q4
68.25 kJ = 3.465 kJ + 0.525 kJ + 0.7875 kJ
e) Comparing the obtained value with the tabulated value, we have: λ = 335 kJ/kg (tabulated value) 68.25 kJ (obtained value)
Since the obtained value (68.25 kJ) is not equal to the tabulated value (335 kJ/kg), we can conclude that there might be some errors or uncertainties in the calculations or assumptions made.
