Answer:
The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg
Explanation:
Heat gain by ice = Heat lost by water
Thus,
Heat of fusion + 
Where, negative sign signifies heat loss
Or,
Heat of fusion + 
Heat of fusion = 334 J/g
Heat of fusion of ice with mass x = 334x J/g
For ice:
Mass = x g
Initial temperature = 0 °C
Final temperature = 6 °C
Specific heat of ice = 1.996 J/g°C
For water:
Volume = 353 mL
Density of water = 1.0 g/mL
So, mass of water = 353 g
Initial temperature = 26 °C
Final temperature = 6 °C
Specific heat of water = 4.186 J/g°C
So,
345.976x = 29553.16
x = 85.4197 kg
Thus,
<u>The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg</u>
Answer:
Explanation:
Firstly of all we have to construct the min-heap of the k-sub list and each sub list which is a node in the constructed min-heap.
We have several steps to follows:
Step-1. When we compare the two sub lists, at the starting we can compare their first elements which is actually their minimum elements.
Step-2. The min-heap formation will cost be O(k) time.
Step-3. After the step 1 & step 2 we can run the minimum algorithm which can be extracted from the minimum element in the root list.
Step-4. Then Update the root list in the heap and after that simplify the min-heap as maintained by the new minimum element in the root list.
Step-5. If any root sub-list becomes empty in the step 4 then we can take any leaf sub-list from the root and simplify it.
Step-6. At every Extraction of the element it can take up to O(log k) time.
Hence, We can say that the extract of n element in the total whose
Running time will be O(n log k + k) which can be equal to the O(n log k+ k) (since k < n).
If two people pull on it, it maintains its original size and shape. Hope it helps.
