Answer:
The mass of water is 25.9g.
Explanation:
1st) It is necessary to use the Heat formula. Knowing that the piece of aluminum releases heat and the water absorbs heat, the equation is a sum of both heats euqal to zero:
[tex]Q_{Al}+Q_{water}=0[/tex]
2nd) The information given in the exercise is:
• Piece of aluminum:
- Mass (mAl)=55.0g
- Heat capacity (cAl)=0.902J/°C*g
- Initial Temperature (TiAl)= 72.4°C
- Final Temperature (TfAl)= 44.9°C
• Water:
- Mass (mwater)= this is what we have to calculate.
- Heat capacity (cwater)= 4.18J/°C*g
- Initial Temperature (Tiwater)= 32.3°C
- Final Temperature (Tfwater)= 44.9°C
3rd) It is necessary to replace the values in the formula to calculate the mass of the aluminum piece:
[tex]\begin{gathered} -Q_{Al}=Q_{water} \\ -\lbrack m_{Al}*c_{Al}*(T_{fAl}-T_{iAl})\rbrack=m_{water}*c_{water}*\left(T_{fwater}-T_{iwater}\right) \\ -\lbrack55.0g*0.902\frac{J}{\degree C*g}*(44.9°C-72.4°C)\rbrack=m_{water}*4.18\frac{J}{\operatorname{\degree}C*g}*(44.9°C-32.3°C) \\ -\lbrack-1364.28J\rbrack=m_{water}*52.67\frac{J}{\operatorname{\degree}C} \\ \frac{1364.28J}{52.67\frac{J}{g}}=m_{water} \\ m_{water}=25.9g \\ \end{gathered}[/tex]
Resolution number 2 (image):
Finally, the mass of water is 25.9g.
