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A 100.0 g sample of titanium at [tex]$63.0^{\circ} C$[/tex] is warmed to [tex]$85.0^{\circ} C$[/tex]. How much energy is needed to warm the metal?

[tex]\[
\begin{aligned}
\left( c _{ Ti }\right. & \left.=0.130 \frac{ J }{ g \cdot{ }^{\circ} C }\right) \\
q & =[?] J
\end{aligned}
\][/tex]


Sagot :

To determine how much energy is needed to warm the titanium, we will use the formula for calculating the amount of heat energy required to change the temperature of a substance:

[tex]\[ q = m \cdot c \cdot \Delta T \][/tex]

where:
- [tex]\( q \)[/tex] is the heat energy in joules (J).
- [tex]\( m \)[/tex] is the mass of the substance in grams (g).
- [tex]\( c \)[/tex] is the specific heat capacity of the substance in joules per gram per degree Celsius [tex]\(\frac{J}{g \cdot {}^\circ C}\)[/tex].
- [tex]\(\Delta T \)[/tex] is the change in temperature in degrees Celsius [tex]\((^\circ C)\)[/tex], calculated as the final temperature minus the initial temperature.

Let's break down the problem into steps:

1. Identify the given values:
- Mass of titanium, [tex]\( m = 100.0 \, \text{g} \)[/tex]
- Specific heat capacity of titanium, [tex]\( c = 0.130 \, \frac{J}{g \cdot {}^\circ C} \)[/tex]
- Initial temperature, [tex]\( T_{\text{initial}} = 63.0^\circ C \)[/tex]
- Final temperature, [tex]\( T_{\text{final}} = 85.0^\circ C \)[/tex]

2. Calculate the change in temperature ([tex]\(\Delta T\)[/tex]):
[tex]\[ \Delta T = T_{\text{final}} - T_{\text{initial}} \][/tex]
[tex]\[ \Delta T = 85.0^\circ C - 63.0^\circ C = 22.0^\circ C \][/tex]

3. Substitute the values into the formula to find [tex]\( q \)[/tex]:
[tex]\[ q = m \cdot c \cdot \Delta T \][/tex]
[tex]\[ q = 100.0 \, \text{g} \cdot 0.130 \, \frac{J}{g \cdot {}^\circ C} \cdot 22.0^\circ C \][/tex]

4. Perform the multiplication:
[tex]\[ q = 100.0 \cdot 0.130 \cdot 22.0 \][/tex]
[tex]\[ q = 286.0 \, \text{J} \][/tex]

So, the amount of energy needed to warm the 100.0 g sample of titanium from [tex]\( 63.0^\circ C \)[/tex] to [tex]\( 85.0^\circ C \)[/tex] is [tex]\( 286.0 \)[/tex] joules.