50 grams of propane is placed into a bomb calorimeter at 20 deg c. it is found that the heat released is 50 kj. what is the temperature of the system? the heat capacity of the cariometer is 6.00 kj/c

The final temperature is 28 °C.

There are two heat transfers involved.

heat from combustion of propane + heat gained by water = 0

q_1 + q_2 = 0

q_1+ C_calΔT = 0

q_1 = -50 kJ

q_2 = 6.00 kJ·°C^(-1)  × ΔT = 6.00 ΔT kJ·°C^(-1)

q_1 + q_2 = -50 kJ + 6.00 ΔT kJ·°C^(-1) = 0

ΔT = 50/[6.00 °C^(-1)] = 8.33 °C

T_f = T_i + ΔT = 20 °C + 8.33 °C = 28 °C

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.  

the transfer of heat by electromagnetic radiation is called thermal radiation . it is usually carried out by electromagnetic waves in an energy range just below the visible spectrum, which is the infrared range.

so the answer would be radiation.

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