Question

Як зміниться температура алюмінієвої заготовки масою 2 кг, якщо їй передати таку саму кількість теплоти, яка витрачається на нагрівання 880 г води від 0 0С до 100 0С? питома теплоємність алюмінію 920 Дж/кг*0С.

Answer 1

                         Q1=ca*ma*дельта t

                      Q2= cв*mв*(t2-t1)

                         Q1=Q2

                    ca*ma*дельта t=cв*mв*(t2-t1)

               дельта t = (св*mв*100)/(ca*ma)

                    дельта t= (4200*0,88*100)/(920*2)

                     дельта t=369600/1840

                     дельта t=200,8 0C

Answer 2

ma=2кг                          Q1=ca*ma*дельта t

mв=0,88кг                      Q2= cв*mв*(t2-t1)

t1=0 0C                         Q1=Q2

t2=100 0C                      ca*ma*дельта t=cв*mв*(t2-t1)

c=920Дж/кг*0С               дельта t = (св*mв*100)/(ca*ma)

дельта t=?                     дельта t= (4200*0,88*100)/(920*2)

                                   дельта t=369600/1840

                                   дельта t=200,8 0C

Answer 3

Calculation of Temperature Change

To calculate the change in temperature of an aluminum ingot with a mass of 2 kg, we can use the formula:

Q = mcΔT

Where: - Q is the amount of heat transferred - m is the mass of the aluminum ingot - c is the specific heat capacity of aluminum - ΔT is the change in temperature

Given that the specific heat capacity of aluminum is 920 J/(kg·°C) and the amount of heat transferred is the same as that required to heat 880 g of water from 0°C to 100°C, we can proceed with the calculation.

First, let's convert the mass of water from grams to kilograms:

m_water = 880 g = 0.88 kg

Next, we can calculate the amount of heat transferred to the water using the formula:

Q_water = mcΔT

Where: - Q_water is the amount of heat transferred to the water - m_water is the mass of the water - c_water is the specific heat capacity of water - ΔT_water is the change in temperature of the water

The specific heat capacity of water is approximately 4,186 J/(kg·°C).

Now, let's calculate the amount of heat transferred to the water:

Q_water = m_water * c_water * ΔT_water

Given that the initial temperature of the water is 0°C and the final temperature is 100°C, the change in temperature of the water is:

ΔT_water = 100°C - 0°C = 100°C

Substituting the values into the formula:

Q_water = 0.88 kg * 4,186 J/(kg·°C) * 100°C = 368,288 J

Since the amount of heat transferred to the aluminum ingot is the same as that transferred to the water, we can equate the two equations:

Q_water = Q_aluminum

m_water * c_water * ΔT_water = m_aluminum * c_aluminum * ΔT_aluminum

Substituting the known values:

0.88 kg * 4,186 J/(kg·°C) * 100°C = 2 kg * 920 J/(kg·°C) * ΔT_aluminum

Simplifying the equation:

ΔT_aluminum = (0.88 kg * 4,186 J/(kg·°C) * 100°C) / (2 kg * 920 J/(kg·°C))

Calculating the value:

ΔT_aluminum ≈ 22.78°C

Therefore, the temperature change of the aluminum ingot with a mass of 2 kg, when the same amount of heat is transferred to it as is transferred to 880 g of water from 0°C to 100°C, is approximately 22.78°C.

Conclusion

The temperature of the aluminum ingot with a mass of 2 kg will increase by approximately 22.78°C when the same amount of heat is transferred to it as is transferred to 880 g of water from 0°C to 100°C.