Answer:
The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .
Step-by-step explanation:
Given as :
The mass of liquid water = 50 g
The initial temperature = 
= 15°c
The final temperature = 
= 100°c
The latent heat of vaporization of water = 2260.0 J/g
Let The amount of heat required to raise temperature = Q Joule
Now, From method
Heat = mass × latent heat × change in temperature
Or, Q = m × s × ΔT
or, Q = m × s × ( - )
So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )
Or, Q = 50 g × 2260.0 J/g × 85°c
∴ Q = 9,605,000 joule
Or, Q = 9,605 × 10³ joule
Or, Q = 9605 kilo joule
Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer
Answer:
15 more
Step-by-step explanation:
John has 36 sweets and he shares them in the ratio 2 : 7.
How many sweets is the larger share?
Answer:
2. 13
Step-by-step explanation:
The least number of classes need for all of the students to be registered in a class is 13 since you want all of the stuednts to be registered so no one should be left out. If we do 12*12 (12 students in each class and there are 12 classes), that would only let 144 students take the swimming classes and 8 students would be left out. We don't want that though since we want all of the students to be registered. So let's go to 12*13 (12 students in each class and there are 13 classes), that would let 156 students take the swimming class. 156>152 so therefore, 13 classes would allow for all 152 students to be registered and it is the least number of classes needed for all the students to be registered in a class (plus you would have 4 seats left for anyone who wants to register in the future but the remainder doesn't matter).
