Question

Получив от горячей воды 100 Дж тепла,алюминиевая ложка массой 0,1 кг нагреется на....°C

Answer 1

Дано:
Q=100Дж
c=920 Дж/кг•°С
m=0,1 кг
Найти:
∆t
Решение:
Q=cm∆t следовательно ∆t=Q/cm
∆t=100Дж/920Дж /кг°С•0,1кг=~1,08°С

Answer 2

Calculating the Temperature Increase of an Aluminum Spoon

To calculate the temperature increase of an aluminum spoon after receiving 100 J of heat, we can use the specific heat capacity formula:

Q = mcΔT

Where: - Q = heat energy (in joules) - m = mass (in kilograms) - c = specific heat capacity (in joules per kilogram per degree Celsius) - ΔT = change in temperature (in degrees Celsius)

We can rearrange the formula to solve for ΔT:

ΔT = Q / mc

Given: - Q = 100 J (heat energy) - m = 0.1 kg (mass of the aluminum spoon)

We need to find the specific heat capacity (c) of aluminum to calculate the temperature increase.

Specific Heat Capacity of Aluminum

The specific heat capacity of aluminum is approximately 900 J/kg°C.

Calculating the Temperature Increase

Now, we can calculate the temperature increase of the aluminum spoon using the specific heat capacity of aluminum:

ΔT = 100 J / (0.1 kg * 900 J/kg°C)

ΔT ≈ 1000°C

Therefore, the aluminum spoon would heat up by approximately 1000°C after receiving 100 J of heat.