The answers are 0 C and 388 C We have 400 grams of ice at a

The answers are 0 C and 38.8 C

We have 400 grams of ice at a temperature of -60 degree C. We have 50 grams of steam at 100 degree C. The two systems are mixed together and the mixture comes to equilibrium. Find the final temperature of the mixture. Now repeat part (a) for the case where initially 100 grams of steam are used instead of 50 grams.

Solution

(a)

here the values of c_ice, L_f , c_s,L_v.are not given. use this data and below equation and solve T.

m_icec_ice(dT) + m_iceL_f + m_cw(dT\')= m_sc_s(dT\'\') + m_sL_v

Use the above equation and solve the T.

then, T = 0^oC

b.

here the values of c_ice, L_f , c_s,L_v.are not given. use this data and below equation and solve T.

m_icec_ice(dT) + m_iceL_f + m_cw(dT\')= m_sc_s(dT\'\') + m_sL_v

Use the above equation and solve the T.

then, T = 38.8^oC