Question

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?

Answer 1

Answer:

a) $T_f=\frac{m_1\lambda_{H20}-m_2c_2T_i}{m_2c_2-m_1c_1}$

b) 295.52K

Explanation:

a) you use the following formula:

$Q_1+Q_2=Q$

$-m_2c_2(T_f-T_{i2})-m\lambda_{H20}=m_1c_1(T_f-T_{i1})$    ( 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):

$T_f=-\frac{m_1\lambda_{H20}-m_1c_1T_{i1}-m_2c_2T_{i2}}{m_2c_2+m_1c_1}$

$T_f=\frac{m_1\lambda_{H20}-m_2c_2T_i}{m_2c_2-m_1c_1}$

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

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

Answer 2

Answer:

a) T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

b) T = 295.37 K

Explanation:

Given;

Initial temperature of tea T1 = 31 C

Initial temperature of ice T2 = 0 C

Mass of tea m1 = 0.89 kg

Mass of ice m2 = 0.075kg

The heat capacity of both water and tea c = 4186 J/(kg⋅K)

the latent heat of fusion for water is Lf = 33.5 × 10^4 J/kg

And T = the final temperature of the mixture

Heat loss by tea = heat gained by ice

m1c∆T1 = m2c∆T2 + m2Lf

m1c(T1-T) = m2c(T-T2) + m2Lf

m1cT1 - m1cT = m2cT - m2cT2 + m2Lf

m1cT + m2cT = m1cT1 + m2cT2 - m2Lf

T(m1c + m2c) = m1cT1 + m2cT2 - m2Lf

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

Substituting the values;

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

T = (0.89×4186×31 + 0.075×4186×0 - 0.075×33.5 × 10^4)/(0.89×4186 + 0.075×4186)

T = 22.37 °C

T = 273 + 22.37 K

T = 295.37 K