# What mass of water will change its temperature by 3 C when 525 J of heat is added to it? The specific heat of water is 4-.8 J/gC

To obtain the mass of water, let's use the equation below:

Based on the information you've provided, we know the following variables:

All we have to do is rearrange the equation to solve for m. This can be accomplished by dividing both sides by

Now, we just plug in the known values:

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To find the mass of water that will change its temperature by 3°C when 525 J of heat is added, you use the formula:

[q = mcΔT]

Where:

- (q) = heat energy (in joules)
- (m) = mass of the substance (in grams)
- (c) = specific heat capacity of the substance (in J/g°C)
- ΔT = change in temperature (in °C)

Rearrange the formula to solve for mass:

[m = \frac{q}{cΔT}]

Substitute the given values:

[m = \frac{525 , \text{J}}{(4.18 , \text{J/g°C}) \times 3 , \text{°C}}]

[m \approx 40 , \text{grams}]

So, the mass of water that will change its temperature by 3°C when 525 J of heat is added is approximately 40 grams.

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