Question

Сколько джоулей тепла выделится при остывании 9 см³ никеля на 27 ºС? Плотность никеля 8,9 г/см³. Удельная теплоёмкость никеля 460 Дж/(кг׺С).

Answer 1

M=ρV
8.9г/см³*9см³=80,1г=0,08кг
Q=сmt°
460Дж/кг*°С*0,08кг*27°С=993Дж 

Answer 2

Answer in detail. How much heat will be released when 9 cm³ of nickel cools by 27 ºC? The density of nickel is 8.9 g/cm³. The specific heat capacity of nickel is 460 J/(kg׺C).

To answer this question, we need to use the formula for heat transfer:

$$Q = mc\Delta T$$

where $Q$ is the amount of heat transferred, $m$ is the mass of the substance, $c$ is the specific heat capacity of the substance, and $\Delta T$ is the change in temperature.

First, we need to find the mass of 9 cm³ of nickel, using the density:

$$m = \rho V = 8.9 \times 9 = 80.1 \text{ g}$$

Then, we need to convert the mass from grams to kilograms, by dividing by 1000:

$$m = 80.1 \div 1000 = 0.0801 \text{ kg}$$

Next, we need to find the change in temperature, which is given by subtracting the final temperature from the initial temperature. Since we are only interested in the magnitude of the change, we can ignore the sign:

$$\Delta T = |T_f - T_i| = |27 - T_i|$$

We do not know the initial temperature $T_i$, but we can assume that it is higher than 27 ºC, since the nickel is cooling down. Therefore, we can write:

$$\Delta T = T_i - 27$$

Finally, we can plug in the values into the formula for heat transfer and calculate the amount of heat released:

$$Q = mc\Delta T = 0.0801 \times 460 \times (T_i - 27)$$

$$Q = 16.766 \times (T_i - 27) \text{ J}$$

This is the general expression for the amount of heat released by 9 cm³ of nickel when it cools by 27 ºC. To find the exact value, we need to know the initial temperature of the nickel. For example, if the initial temperature was 100 ºC, then the amount of heat released would be:

$$Q = 16.766 \times (100 - 27) \text{ J}$$

$$Q = 1223.498 \text{ J}$$

I hope this answer helps you understand the concept of heat transfer and how to apply it to different situations.