At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

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.