Question

На сколько изменится температура воды объемом 300 мл если она получит Всю энергию выделившуюся при охлаждении свинцового цилиндра массой 0,21 кг от температуры 120 градусов до температуры 20 градусов

Answer 1

Дельта t= Q/m воды * c воды
м воды = pV
m воды = 1000 кг/м3 * 0,0003 м3= 0,3 кг
Q= m цилиндра * с цилиндра * (t1-t2)
Q = 0,21 кг * 140 Дж/кг*С * (120 С - 20 С)= 2940 Дж
дельта t = 2940 Дж/ 0,3 кг* 4200 Дж/кг*С~2 С
Ответ: на 2 градуса С

Answer 2

Calculation of Temperature Change

To calculate the change in temperature of the water, we can use the formula:

Q = mcΔT

Where: - Q is the heat energy transferred - m is the mass of the water - c is the specific heat capacity of water - ΔT is the change in temperature

Given: - Mass of the water (m) = 300 ml = 0.3 kg - Initial temperature (T1) = 120 degrees Celsius - Final temperature (T2) = 20 degrees Celsius - Specific heat capacity of water (c) = 4.18 J/g°C

First, we need to convert the mass of water from grams to kilograms: - Mass of the water (m) = 300 ml = 300 g = 0.3 kg

Next, we can calculate the change in temperature (ΔT): - ΔT = T2 - T1 = 20°C - 120°C = -100°C

Now, we can calculate the heat energy transferred (Q): - Q = mcΔT = 0.3 kg * 4.18 J/g°C * -100°C = -125.4 J

Therefore, the change in temperature of the water will be -100°C and the heat energy transferred will be -125.4 J.