A sample of aluminum with mass of 53.3 g is initially at 143 ∘C

What is the temperature of the aluminum after it loses 2.76 x 103 J?

Well, losing all that energy must have made the aluminum pretty cool! In fact, it's enough to bring down the temperature quite a bit.

To find the new temperature, we'll need to use the specific heat capacity of aluminum, which is approximately 0.897 J/g°C.

First, we need to find out how much the temperature changes. To do that, we can use the equation:

Q = m * c * ΔT

Where:
Q is the amount of energy lost (2.76 x 10^3 J)
m is the mass of the aluminum (53.3 g)
c is the specific heat capacity of aluminum (0.897 J/g°C)
ΔT is the change in temperature (unknown)

Rearranging the equation, we get:

ΔT = Q / (m * c)

Now we can substitute in the values:

ΔT = 2.76 x 10^3 J / (53.3 g * 0.897 J/g°C)

Calculating that, we find:

ΔT ≈ 0.0613 °C

Since we started at 143 °C and the temperature decreased, we need to subtract the change from the initial temperature:

New Temperature = 143 °C - 0.0613 °C

Therefore, the temperature of the aluminum after losing 2.76 x 10^3 J is approximately 142.9387 °C.

To determine the final temperature of the aluminum, we can use the specific heat formula:

Q = mcΔT

Where:
Q = Heat energy transferred (in Joules)
m = Mass of the aluminum (in grams)
c = Specific heat capacity of aluminum (in J/g∙°C)
ΔT = Change in temperature (final temperature - initial temperature, in °C)

Given:
Initial mass (m) = 53.3 g
Initial temperature (initial T) = 143 °C
Heat energy transferred (Q) = 2.76 x 10^3 J

First, we need to convert the mass from grams to kilograms since specific heat capacity is usually given in J/g∙°C:

m = 53.3 g / 1000 = 0.0533 kg

Next, we need to determine the specific heat capacity of aluminum. The specific heat capacity of aluminum is typically 0.897 J/g∙°C or 897 J/kg∙°C. Here we will use the value of 0.897 J/g∙°C.

Q = mcΔT
ΔT = Q / (mc)

ΔT = (2.76 x 10^3 J) / (0.0533 kg * 0.897 J/g∙°C) = 60.96 °C

Finally, we can find the final temperature by subtracting the change in temperature (ΔT) from the initial temperature (initial T):

Final temperature (final T) = initial T - ΔT
Final temperature (final T) = 143 °C - 60.96 °C = 82.04 °C

Therefore, the temperature of the aluminum after it loses 2.76 x 10^3 J is approximately 82.04 °C.

To find the final temperature of the aluminum after it loses energy, we can use the equation:

Q = mcΔT

Where:
Q is the heat energy lost by the aluminum
m is the mass of the aluminum
c is the specific heat capacity of aluminum
ΔT is the change in temperature

First, let's calculate the heat energy lost by the aluminum using the given information:

Q = 2.76 x 10^3 J (given)

Next, we need to know the specific heat capacity of aluminum. The specific heat capacity of aluminum is approximately 0.897 J/g·°C.

Now, let's rearrange the equation to solve for the change in temperature (ΔT):

ΔT = Q / (mc)

Substituting the given values:

ΔT = (2.76 x 10^3 J) / (53.3 g * 0.897 J/g·°C)

ΔT = (2.76 x 10^3 J) / (47.7691 J·°C/g)

ΔT ≈ 57.7 °C

Finally, to find the final temperature, we need to subtract the change in temperature from the initial temperature:

Final temperature = Initial temperature - ΔT

Final temperature = 143 °C - 57.7 °C

Final temperature ≈ 85.3 °C

Therefore, the temperature of the aluminum after it loses 2.76 x 103 J is approximately 85.3 °C.

heat=mass*c*(Tf-Ti)

-2.75J=53.3g*specific heat*(Tf-143C)
specific heat .9J/gC
Tf=143-2.76e3/(.9*53.3)=85.5 C
check all that logic and math.