A 2.8 kg gold bar at 99 degrees celcius
is dropped into 0.21 kg of water at 23 degrees celcius
What is the final temperature? Assume the
specific heat of gold is 129 J/kg times degrees celcius. Answer in units of degrees celcius
To find the final temperature when the gold bar is dropped into water, we can use the principle of heat transfer. The heat lost by the gold bar will be equal to the heat gained by the water.
The formula for heat transfer is:
q = m * c * ΔT
where:
q is the heat transfer (in joules),
m is the mass of the substance (in kg),
c is the specific heat capacity (in J/kg·°C),
ΔT is the change in temperature (in °C).
In this case, the heat lost by the gold bar is equal to the heat gained by the water. We can set up the equation as follows:
q (gold) = q (water)
(m (gold) * c (gold) * ΔT (gold)) = (m (water) * c (water) * ΔT (water))
Given:
m (gold) = 2.8 kg,
c (gold) = 129 J/kg·°C,
ΔT (gold) = final temperature - initial temperature = final temperature - 99°C,
m (water) = 0.21 kg,
c (water) = specific heat of water = 4186 J/kg·°C,
ΔT (water) = final temperature - initial temperature = final temperature - 23°C.
Substituting the given values into the equation:
(2.8 kg * 129 J/kg·°C * (final temperature - 99°C)) = (0.21 kg * 4186 J/kg·°C * (final temperature - 23°C))
Now we can solve for the final temperature.
To simplify the equation, let's distribute the multiplication on both sides:
(361.2 * final temperature - 98,628) = (879.06 * final temperature - 19,244.38)
Let's move all terms with the final temperature to one side and all constant terms to the other side:
361.2 * final temperature - 879.06 * final temperature = -98,628 + 19,244.38
-517.86 * final temperature = -79,383.62
Dividing both sides of the equation by -517.86:
final temperature = -79,383.62 / -517.86
final temperature ≈ 153.15°C
Therefore, the final temperature is approximately 153.15°C.