A200 g sample of water with a temperature of 32◦c is added to 50 g water at 55◦c in an insulated container. what is the final temperature after thermal equilibrium is reached? the specific heat of water in the liquid state is 4.18 j/g◦c, in the solid state is 2.09 j/g◦c, and in the gaseous state is 2.03 j/g◦c. the heat of fusion is 334 j/g and the heat of vaporization is 2260 j/g. 1. 49.2◦c 2. 36.6◦c 3. 52.6◦c 4. 38.4◦c 5. 50.4◦c 6. 43.5◦c 7. 34.1◦c

2. 36.6°C.

Explanation:

∵ The amount of heat released by hot water = amount of heat absorbed by cold water.

•The amount of heat released by hot water = Q hot water = - m.c.ΔT.

where, m is the mass of hot water = 50 g.

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = final T - 55°C).

•The amount of heat absorbed by cold water = Q cold water = m.c.ΔT.

where, m is the mass of cold water = 200 g.

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = final T - 32°C).

∵ Q hot water = Q cold water  

∴ (- m.c.ΔT) hot water = (m.c.ΔT)cold water

∴ (- m.ΔT) hot water = (m.ΔT)cold water

- (50 g)(final T - 55°C) = (200 g)(final T - 32°C)

-50 final T + 2750 = 200 final T - 6400  

250 final T = 2750 + 6400 = 9150

∴ final T = 9150/250 = 36.6°C.

the answer is alpha particles hope i you

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

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