Answer:
The specific heat of gold is 0.129 J/g°C
Explanation:
Step 1: Data given
Mass of gold = 15.3 grams
Heat absorbed = 87.2 J
Initial temperature = 35.0 °C
Final temperature = 79.2 °C
Step 2:
Q = m*c*ΔT
⇒ Q =the heat absorbed = 87.2 J
⇒ m = the mass of gold = 15.3 grams
⇒ c = the specific heat of gold = TO BE DETERMINED
⇒ ΔT = The change in temperature = T2 - T1 = 79.2 - 35.0 = 44.2 °C
87.2 J = 15.3g * c * 44.2°C
c = 87.2 / (15.3 * 44.2)
c = 0.129 J/g°C
The specific heat of gold is 0.129 J/g°C
Answer:
<u>225.6 kJ</u>, <em>assuming the water is already at 100 °C</em>
Explanation:
The correct answer to this question will depend on the initial temperature of the water to which heat is added to produce steam. Energy is required to raise the water temperature to 100°C. At that point, an energy of vaporization is needed to convert liquid water at 100 °C to water vapor at 100°C. The heat of vaporization for water is 2256.4 kJ/kg. The energy required to bring 100g of water from a lower temperature to 100°C is calculated at 4.186 J/g°C. We don't know the starting temperature, so this step cannot be calculated.
<em><u>Assuming</u></em> that we are already at 100 °C, we can calculate the heat required for vaporization:
(100.0g)(1000.0g/1 kg)(2256.4 kJ/kg) = 225.6 kJ for 100 grams water.
Explanation:
they are white so they can camouflage, but actually the polar bears skin is black and it's hair is hollow
Answer:
Va = (MbVb)/Ma
Explanation:
Divide both sides by Ma and voila!
Answer:
Explanation:
Hello there!
In this case, according to the given information, it will be possible for us to solve this problem by using the Boyle's law as an inversely proportional relationship between pressure and volume:
In such a way, we solve for the final volume, V2, and plug in the initial volume and pressure and final pressure to obtain:
Regards!
