Answered

A 10.35g piece of copper metal was heated to 85.5 degrees C and then placed into 23.6g of water. the initial temperate of the water was 18.3 degrees C. what was the final temperature?



Answer :

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

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]

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