Answer:
An insulated beaker with negligible mass contains liquid water with a mass of 0.205kg and a temperature of 79.9 °C How much ice at a temperature of −17.5 °C must be dropped into the water so that the final temperature of the system will be 31.0 °C? Take the specific heat for liquid water to be 4190J/Kg.K, the specific heat for ice to be 2100J/Kg.K, and the heat of fusion for water to be 334000J/kg.
The answer to the above question is
Therefore 0.1133 kg ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C
Explanation:
To solve this we proceed by finding the heat reaquired to raise the temperature of the water to 31.0 C from 79.9 C then we use tht to calculate for the mass of ice as follows
ΔH = m×c×ΔT
= 0.205×4190×(79.9 -31.0) = 42002.655 J
Therefore fore the ice, we have
Total heat = mi×L + mi×ci×ΔTi = mi×334000 + mi × 2100 × (0 -−17.5) = 42002.655 J
370750×mi = 42002.655 J
or mi = 0.1133 kg
Therefore 0.1133 kg ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C
<u>Given:</u>
The number of particles = 3.131 * 10²⁴
<u>To determine:</u>
The number of moles corresponding to the given particles
<u>Explanation:</u>
1 mole corresponds to Avogadro's number of particles = 6.023 * 10²³ particles
Therefore, the given particles would correspond to:
3.131* 10²⁴ particles * 1 moles/6.023 * 10²³ particles = 5.199 moles
Ans: A) 5.199 moles are contained in 3.131* 10²⁴ particles
I think it's option C and option D but if you can only choose one answer then I think it's option C