# How do you determine the mass of a water sample that is heated from an initial temperature of 25 deg C to a final temperature of 100 deg C following the addition of 1200 J of heat energy?

Here's how I would do it.

The formula for the amount of heat absorbed by a substance is

where

You can rearrange the formula to calculate the mass:

In your problem,

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You can determine the mass of the water sample using the formula:

( q = mcΔT )

Where:

- ( q ) is the heat energy (in joules)
- ( m ) is the mass of the water (in grams)
- ( c ) is the specific heat capacity of water (4.18 J/g°C)
- ( ΔT ) is the change in temperature (in °C)

First, calculate the change in temperature:

( ΔT = T_f - T_i ) ( ΔT = 100°C - 25°C = 75°C )

Then, rearrange the formula to solve for mass:

( m = \frac{q}{cΔT} )

Substitute the given values:

( m = \frac{1200 J}{4.18 J/g°C * 75°C} ) ( m ≈ 4.83 g )

So, the mass of the water sample is approximately 4.83 grams.

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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.

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.

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.

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