ASAP PLEASE: A man makes a cup of tea using 354 g of water at 93 degrees C inside a 92 g ceramic mug with a specific heat of 1.09 J/g degrees C. He cools the tea by adding an ice cube at -6 degrees C. The final temp of the tea and melted ice together is 72 degrees C. Assume the mug and tea keep the same temp. Calculate the mass of the ice cube.

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward