Question

example: a 125.0 g sample of aluminum pellets at 215.0°c is added to 45.0 g of water at 22.0°c. what is the final temperature in celcius of the water? (assume no energy goes into or comes out of the container.) enter only the number with no units.

Explanation:

Step1: Write the heat - transfer formula

According to the principle of conservation of energy \(q_{lost}=q_{gained}\). The heat - transfer formula is \(q = mc\Delta T\), where \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T=T_{final}-T_{initial}\). For aluminum, \(m_{Al}=125.0\ g\), \(c_{Al}=0.900\ J/g^{\circ}C\), \(T_{i,Al}=215.0^{\circ}C\). For water, \(m_{w}=45.0\ g\), \(c_{w}=4.184\ J/g^{\circ}C\), \(T_{i,w}=22.0^{\circ}C\). Let the final temperature be \(T_f\). Then \(-m_{Al}c_{Al}(T_f - T_{i,Al})=m_{w}c_{w}(T_f - T_{i,w})\)

Step2: Substitute the values into the formula

\(-125.0\times0.900\times(T_f - 215.0)=45.0\times4.184\times(T_f - 22.0)\)

First, expand both sides:
\(-112.5(T_f - 215.0)=188.28(T_f - 22.0)\)

\(-112.5T_f+112.5\times215.0 = 188.28T_f-188.28\times22.0\)

\(-112.5T_f + 24187.5=188.28T_f-4142.16\)

Step3: Rearrange the equation to solve for \(T_f\)

\(-112.5T_f-188.28T_f=-4142.16 - 24187.5\)

\((-112.5 - 188.28)T_f=-28329.66\)

\(-300.78T_f=-28329.66\)

\(T_f=\frac{28329.66}{300.78}\)

\(T_f = 94.2\)

Answer:

\(94.2\)