Ablock of ice at 0°c is added to a 150g aluminum calorimeter cup which holds 210 g of water at 12°c. if all but 2.0 g of ice melt, what was the original mass of the block of ice?

original mass of the block of ice is 38.34 gram

Explanation:

Given data

cup mass = 150 g

ice temperature = 0°C

water mass = 210 g

water temperature = 12°C

ice melt = 2 gram

to find out

solution

we know here

specific heat of aluminum is c = 0.900 joule/gram °C

Specific heat of water C =  4.186 joule/gram °C

so here temperature difference is dt =  12- 0 = 12°C

so here heat lost by water and cup are given by

heat lost  = cup mass × c  × dt + water  mass × C × dt

heat lost  = 150 × 0.900  × 12 + 210 × 4.186 × 12

heat lost  = 12168.72 J

so

mass of ice melt here = heat lost / latent heat of fusion

here we know latent heat of fusion = 334.88 joule/gram

so

mass of ice melt  =  12168.72 / 334.88

mass of ice melt  is 36.337554 gram

so mass of ice is here = mass of ice melt + ice melt

mass of ice  =  36.337554 + 2

mass of ice  =  38.337554 gram

so original mass of the block of ice is 38.34 gram

c) up is the direction the magnetic force is acting on the charge.

im srry but ur wrong the answer is d a smooth rock, some red dirt, and an empty bird's nest.