How many calories of are required to heat 731 grams of water from 35°c to 83°c? (assume that the specific heat of water is 1.00
2 answers:
Answer:
The calories required = 35088
Explanation:
The heat required can be calculated by using specific heat of the water and mass and temperature raised.
The specific heat of water is the amount of heat required to raise the temperature of one gram of water by one degree celsius
Thus
Heat required = mass of water X specific heat X change in temperature
Given:
Specific heat of water = 1 cal /g °C
mass of water = 731 g
change in temperature = 83-35 = 48°C
Putting values
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Explanation:
1)
Mass of NaOH = m
MOlar mass of NaOH = 40 g/mol
Volume of NaOH solution = 1.00 L
Molarity of the solution= 1.00 M
A student can prepare the solution by dissolving the 40. grams of NaOH in is small volume of water and making that whole volume of solution to volume of 1 L.
Upto two significant figures mass should be determined.
2)
(dilution equation)
Molarity of the NaOH solution =
Volume of the solution =
Molarity of the NaOH solution after dilution =
Volume of NaOH solution after dilution=
A student can prepare NaOH solution of 1.00 M by diluting the 0.500 L of 2.00 M solution of NaOH with water to 1.00 L volume.
Upto three significant figures volume should be determined.