AP EAMCET202018 Sep 2020Evening ShiftPhysicsThermal Properties of MatterActual
(300 ~g ) of water at (25^ C ) is added to (100 ~g ) of ice at (0^ C ). The final temperature of mixture will be
Options
- A(25^ C )
- B(0^ C )
- C(12.5^ C )
- D(30^ C )
Correct answer
B. (0^ C )
Step-by-step solution
Given, mass of water (m_w=300 ~g ) Temperature of water, (T_w=25^ C ) Mass of ice, ( gathered m_i=100 ~g T_i=0^ C gathered ) Heat required to convert the ice at (0^ C ) to water at (0^ C ) ( array lll Q₁ & =m_i L_i=100 80 [L_i=80 cal / g ] & =8000 cal ( i ) array ) Heat given by water from (25^ C ) water to (0^ C ) water ( aligned Q₂ & =m_w c T=300 1 (25-0) & =7500 cal (ii) aligned ) From Eqs. (i) and (ii), we observed that (Q₂ < Q₁ ) Hence, total ice will not be melt, so final temperature of mixture will be (0^ C