Question

Алюминиевую ложку массой 50 г при температуре 20 градусов опускают в горячую воду при температуре 70 градусов какое количество теплоты получит ложка?

Answer 1

Q=cm(t1-t)
50 г=0.05 кг
Q= 920 * 0.05(70-20)=2300 Дж=2.3 кДж

Answer 2

Calculation of Heat Transfer

To calculate the amount of heat gained by the aluminum spoon when it is submerged in hot water, we can use the formula:

Q = m * c * ΔT

Where: - Q is the heat gained or lost by the object (in joules) - m is the mass of the object (in kilograms) - c is the specific heat capacity of the material (in joules per kilogram per degree Celsius) - ΔT is the change in temperature (in degrees Celsius)

In this case, the mass of the aluminum spoon is given as 50 grams, which is equivalent to 0.05 kilograms. The initial temperature of the spoon is 20 degrees Celsius, and it is submerged in hot water at a temperature of 70 degrees Celsius.

Now, let's calculate the heat gained by the spoon.

Calculation Steps:

1. Convert the mass of the spoon to kilograms: - 50 grams = 0.05 kilograms

2. Calculate the change in temperature: - ΔT = final temperature - initial temperature - ΔT = 70 degrees Celsius - 20 degrees Celsius - ΔT = 50 degrees Celsius

3. Determine the specific heat capacity of aluminum: - According to the specific heat capacity of aluminum is approximately 4200 joules per kilogram per degree Celsius.

4. Calculate the heat gained by the spoon using the formula: - Q = m * c * ΔT - Q = 0.05 kg * 4200 J/(kg·°C) * 50 °C

Now, let's calculate the heat gained by the spoon using the given values.

Calculation:

Q = 0.05 kg * 4200 J/(kg·°C) * 50 °C

Q = 10500 J

Therefore, the aluminum spoon will gain 10500 joules of heat when submerged in the hot water at 70 degrees Celsius.

Please note that this calculation assumes that there is no heat loss to the surroundings and that the specific heat capacity of aluminum remains constant over the given temperature range.