Respuesta :
The amount of heat that is gained by water is equal to the amount of heat loss by the ice. Therefore, we have the following relationship:
[tex]-Q_w=Q_i[/tex]Where "Qw" is the heat of water and "Qi" is the heat of ice. The heat of water is given by:
[tex]Q_w=m_wC_w(T_f-T_{0w})[/tex]The amount of heat of ice must be calculated for the two states, solid and liquid. For the solid-state the temperature will be from -30 degrees to 0 degrees, therefore, we have:
[tex]Q_i=m_iC_i(T_f-T_{0i})[/tex]Where Ci is the specific heat of ice in solid-state and is equal to:
[tex]C_i=2090\frac{kJ}{\operatorname{kg}}[/tex]Replacing the values:
[tex]Q_i=(0.05\operatorname{kg})(2090\frac{kJ}{\operatorname{kg}K})(0-(-30C))[/tex]Solving the operations we get:
[tex]Q_i=3135J[/tex]Now we need to determine the amount of heat that needs the ice to convert into liquid. This is given by:
[tex]Q_{i-l}=m_iL_f[/tex]Lf is the latent heat of ice and is equal to:
[tex]L_f=334\frac{kJ}{\operatorname{kg}}[/tex]Replacing the values we get:
[tex]Q_{i-l}=(0.05\operatorname{kg})(334\frac{kJ}{\operatorname{kg}})[/tex]Solving the operations:
[tex]Q_{i-l}=16.7kJ=16700J[/tex]Now we need the amount of heat of liquid ice to its final temperature, this is given by:
[tex]Q_l=m_iC_l(T_f-0)[/tex]Applying the relationship:
[tex]-m_wC_w(T_f-T_{0w})=m_iC_i(T_f-T_{0i})+m_iL_f+m_iC_w(T_f-0)[/tex]Cw is the specific heat of water and is equal to:
[tex]C_w=4184\frac{J}{\operatorname{kg}K}[/tex]And the specific heat of ice is:
[tex]C_i=4184\frac{J}{\operatorname{kg}K}[/tex]Replacing the values. The first two terms on the right side we already calculated and the final temperature is the same for both:
[tex]-(0.4\operatorname{kg})(4184\frac{kJ}{\operatorname{kg}K})(T_f-35)=3135J+16700J+(0.05)(4184\frac{kJ}{\operatorname{kg}K})(T_f)[/tex]Solving operations:
[tex]-1673.6\frac{kJ}{K}(T_f-35)=19835J+209.2\frac{kJ}{K}T_f[/tex]Now we solve for the final temperature:
[tex]-1673.6T_f+58576=19835+209.2T_f[/tex]Subtracting 19835 to both sideS:
[tex]\begin{gathered} -1673.6T_f+58576-19835=209.2T_f \\ -1673.6T_f+38741=209.2T_f \end{gathered}[/tex]Now we add 1673Tf to both sides:
[tex]\begin{gathered} 38741=209.2T_f+1673.6T_f \\ 38741=1882.8T_f \end{gathered}[/tex]Now we divide both sides by 1882.2
[tex]\frac{38741}{1882.8}=T_f[/tex][tex]20.57=T_f[/tex]Therefore, the final temperature is 20.57 °C. This value can be converted into Kelvin using the following relationship:
[tex]T_K=T_C+273.15[/tex]Replacing the temperature:
[tex]T_K=20.57+273.15=293.72[/tex]Therefore, the final temperature is 293.72 K.
