contestada

A thermos contains m1 = 0.89 kg of tea at T1 = 31° C. Ice (m2 = 0.075 kg, T2 = 0° C) is added to it. The heat capacity of both water and tea is c = 4186 J/(kg⋅K), and the latent heat of fusion for water is Lf = 33.5 × 104 J/kg. show answer No Attempt 50% Part (a) Input an expression for the final temperature after the ice has melted and the system has reached thermal equilibrium.Part (b) What is the final temperature in Kelvin?

Respuesta :

mavila18

Answer:

a) [tex]T_f=\frac{m_1\lambda_{H20}-m_2c_2T_i}{m_2c_2-m_1c_1}[/tex]

b) 295.52K

Explanation:

a) you use the following formula:

[tex]Q_1+Q_2=Q[/tex]

[tex]-m_2c_2(T_f-T_{i2})-m\lambda_{H20}=m_1c_1(T_f-T_{i1})[/tex]    ( 1 )

m2: 0.075kg

m1: 0.89kg

Lf= 33.5 × 104 J/kg

Ti2= 0C=273.15K

Ti1 = 31C = 304.15K

Then, you obtain T from the formula (1):

[tex]T_f=-\frac{m_1\lambda_{H20}-m_1c_1T_{i1}-m_2c_2T_{i2}}{m_2c_2+m_1c_1}[/tex]

[tex]T_f=\frac{m_1\lambda_{H20}-m_2c_2T_i}{m_2c_2-m_1c_1}[/tex]

b) By replacing the values of all parameters you obtain:

[tex]T_f=-\frac{(0.075kg)(33.5*10^4J/kg)-(0.89kg)(4186J/kg.K)(304.15K)-(0.075kg)(4186J/kg.K)(273.15K)}{(0.89kg)(4186J/kg.K)+(0.075kg)(4186J/kg.K)}\\\\T_f=295.52K[/tex]

Otras preguntas