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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Answer:
Explanation:
From the given information:
The accuracy depends on the internal diameter of the cylinder. The cylinder with the least internal diameter is obviously more precise.
Let's assume 1% is the error of measurement.
Then, to measure 70 mL from 10 mL cylinder
The error =
= 0.7 mL
However; for a 100 mL cylinder, the error = 1 mL
Now,
The total volume for 10 mL = (70 + 0.7) = 70.7 mL
The total volume for 100 mL = (70 + 1 ) = 71 mL
Suppose the density (d) is same for both
Then;
the mass of 10 mL = ( d × 70.7) g
the mass pf 100 mL = (d × 71) g
Thus, the mass of 100 mL is greater than that of 10 mL.