calories to heat 9.4 g of water from 15 ∘C to 31 ∘C. Express your answer to two significant figures and include the appropriate units.
q1 = mass H2O x specific heat H2O x (Tfinal - Tinitial)
Plug and chug. Post your work if you get stuck.
@ Dr Bob222
so it would be..
9.4g x 4.184 x (31-15)? and also what would the unit be?
9.4 g x 4.184 J/g*C x (31 C - 15 C) = 629.3 g(J/g*C)(C) = J Note how I've written out the units to show the g and C units cancel to leave just J.
The problem asks for 2 s.f. so I would write 630 J. You may prefer to write it as 6.3E2 J but 630 J is correct to 2 s.f. also.
To calculate the calories required to heat a substance, we can use the formula:
q = m * c * ∆T
where:
q = heat energy in calories
m = mass of the substance in grams
c = specific heat capacity of the substance in calories per gram-degree Celsius
∆T = change in temperature in degrees Celsius
Given:
m = 9.4 g (mass of water)
c = 1 calorie/g·°C (specific heat capacity of water)
∆T = 31°C - 15°C = 16°C (change in temperature)
Now, substitute the given values into the formula:
q = 9.4 g * 1 calorie/g·°C * 16°C
Calculate the value:
q = 150.4 calories
Therefore, it requires 150.4 calories to heat 9.4 g of water from 15°C to 31°C. The answer is rounded to two significant figures, and the unit is "calories."