If 213 J raises a sample of water 8.2°C, what is the mass of the water?

Answer 1

6.2 g

Use the formula #q=mCΔT#

q = heat absorbed or released, in joules (J) m = mass C = specific heat capacity (for water, it is #4.18 J / (g * K)#) ΔT = change in temperature

Plug known values into the formula.

#213 = m(4.18)(8.2)# #213 = 34.276m# #6.214260707 = m#

The answer, expressed in 2 significant figures, is 6.2 g.

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Answer 2

Peyton Adams

The mass of the water can be calculated using the formula:

(q = mcΔT)

Where:

(q) is the heat energy absorbed by the water (213 J)
(m) is the mass of the water (unknown)
(c) is the specific heat capacity of water (4.18 J/g°C)
(ΔT) is the change in temperature (8.2°C)

Rearranging the formula to solve for (m):

(m = \frac{q}{cΔT})

Substituting the given values:

(m = \frac{213 , \text{J}}{4.18 , \text{J/g°C} \times 8.2 , \text{°C}})

(m ≈ 6.06 , \text{g})

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.