Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
We know that the heat that escapes the copper metal is equal to the heat that is added to the water bath.
Thus we can use the MC∆T formula to solve for the final temperature
We need to set up the formula for both the change in temperature of the copper as well as the water
thus
Mass(Cu) * Specific Heat capacity(Cu) * (Final temp - initial temp)(Cu) =
Mass(H2O) * Specific Heat of Water(H2O) * (Final - initial temp)(H2O)
Now we can substitute the given info.
specific heat of copper = 0.386 and water is 4.184
10.35g * 0.386 * (x - 85.5) = 23.6g * 4.184 * (x - 18.3)
Now You just need to solve for X. Also remember to take the absolute value of the left side as it will become negative but really the change in heat or ∆H is equal
Thus we can use the MC∆T formula to solve for the final temperature
We need to set up the formula for both the change in temperature of the copper as well as the water
thus
Mass(Cu) * Specific Heat capacity(Cu) * (Final temp - initial temp)(Cu) =
Mass(H2O) * Specific Heat of Water(H2O) * (Final - initial temp)(H2O)
Now we can substitute the given info.
specific heat of copper = 0.386 and water is 4.184
10.35g * 0.386 * (x - 85.5) = 23.6g * 4.184 * (x - 18.3)
Now You just need to solve for X. Also remember to take the absolute value of the left side as it will become negative but really the change in heat or ∆H is equal
Answer:
[tex]T_F=20.91^{\circ}C[/tex]
Explanation:
For this problem we have to start with the heat equation:
[tex]Q=mCp[/tex]ΔT
In which the delta symbol is:
ΔT=[tex]T_f_i_n_a_l-T_i_n_i_t_i_a_l[/tex]
All the heat from copper would be transfer into the water, so:
[tex]Q_m_e_t_a_l=-~Q_w_a_t_e_r[/tex]
Now, we can identify the variables:
[tex]m~_C_u=10.35~g[/tex]
[tex]m~_H_2_O=23.6~g[/tex]
[tex]T_i~_C_u=85.5^{\circ}C[/tex]
[tex]T_i~H_2O=18.3^{\circ}C[/tex]
[tex]Cp~_C_u~=0.385\frac{J}{g^{\circ}C}[/tex]
[tex]Cp~_H_2_O~=4.184\frac{J}{g^{\circ}C}[/tex]
When we put all together in the equation we will have:
[tex](10.35~g)*(0.385\frac{J}{g^{\circ}C})*(T_F-85.5^{\circ}C)=-(23.6~g)*(4.184\frac{J}{g^{\circ}C})*(T_F-18.3^{\circ}C)[/tex]
[tex]3.99~T_F-~340=-98.74~T_F+1806.99[/tex]
[tex]3.99~T_F+~98.74~T_F=340+1806.99[/tex]
[tex]102.72~T_F=2147.68[/tex]
[tex]T_F=20.91^{\circ}C[/tex]
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
