A mixture containing 21.4g of ice (at exactly 0 C) and 75.3g of water (at 55.3 C) is placed in an insulated container. Assuming no heat is lost to the surroundings, what is the final temperature of the mixture?

Question

ookawaiioo · Answer

Correct me if im wrong guys. 

M1T1 + M2T2 = M3T3

M= mass
T = temperature

$$\frac{ M1T1+M2T2 }{ M3 } = T3$$

anonymous · Answer

$$\color{green}{-m_1c_l(T_f - T_i) = m_2l_f + m_2 c_l (T_f - T_i) \ -(75.3)c_l(T_f - 55.3) = (21.4)l_f + (21.4)c_l (T_f - 0) }$$

anonymous · Answer

Use the principle of heat exchange. Heat lost by the warm water = heat gained to melt the ice + heat gained to raise the temperature of the melted ice. Formula is shown above. Plug in the latent heat of fusion for ice (lf) and the specific heat capacity of water (cl). Look them up. Solve for Tf. Convert units if necessary ...