A 250 g sample of water with an initial temperature of 98.8 c loses 7,500 joules of heat. What is the final temperature of the water ?

To calculate the final temperature of the water, you can use the equation:

\(Q = mcΔT\)

Where:
\(Q\) is the heat lost or gained by the water (in joules),
\(m\) is the mass of the water (in grams),
\(c\) is the specific heat capacity of water (4.18 J/g·°C), and
\(\Delta T\) is the change in temperature (final temperature - initial temperature).

Let's calculate it step-by-step:

Step 1: Convert the mass from grams to kilograms
\(m = 250 \, \text{g} = 0.25 \, \text{kg}\)

Step 2: Convert the temperature from degrees Celsius to Kelvin
\(T_i = 98.8 \, \text{°C} + 273.15 = 372.95 \, \text{K}\)

Step 3: Calculate the change in temperature
\(Q = mcΔT\)
\(\Delta T = \frac{Q}{mc}\)
\(\Delta T = \frac{-7500 \, \text{J}}{(0.25 \, \text{kg})(4.18 \, \text{J/g·°C})}\)

Step 4: Calculate the final temperature
\(T_f = T_i - \Delta T\)

Now, let's substitute the values and calculate it:

\(\Delta T = \frac{-7500 \, \text{J}}{(0.25 \, \text{kg})(4.18 \, \text{J/g·°C})} \approx -7155.98 \, \text{°C}\)

\(T_f = 372.95 \, \text{K} - (-7155.98 \, \text{°C})\)

Finally, convert the result back to degrees Celsius:
\(T_f = -6783.03 \, \text{°C}\)

Therefore, the final temperature of the water is approximately -6783.03°C.

To find the final temperature of the water, we need to apply the principle of energy conservation, specifically the formula for heat transfer:

Q = mcΔT

where Q is the amount of heat transferred, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

In this case, the mass of water (m) is given as 250 g, the initial temperature (T1) is given as 98.8 °C, and the heat transfer (Q) is given as 7,500 J. We need to find the final temperature (T2).

First, we need to rearrange the formula to solve for ΔT:

ΔT = Q / (mc)

Next, substitute the given values:

ΔT = 7,500 J / (250 g * c)

The specific heat capacity (c) of water is approximately 4.18 J/g°C.

ΔT = 7,500 J / (250 g * 4.18 J/g°C)

Now, we can calculate ΔT:

ΔT ≈ 7.18 °C

Finally, to find the final temperature (T2), we add ΔT to the initial temperature (T1):

T2 = T1 + ΔT

T2 = 98.8 °C + 7.18 °C

T2 ≈ 106.98 °C

Therefore, the final temperature of the water after losing 7,500 joules of heat is approximately 106.98 °C.

q = mass x specific heat x (Tfinal-Tionitial)

q = -7500. Substitute and solve for Tfinal

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