Question

250g of water at 30degree celcius us contained in a coppwr vessal of mass 50g.
Calculate the mass of ice requird to bring down the temperature of the vessal and its contents to 5 degree celcius

Answer 1

Heat loss by copper vessel when temperature goes to 5C = msdT

= 50 × 0.4 ( 30 - 5 )

= 500 j

Heat loss by water when temperature goes to 5 C

= msdT = 250 × 4.2 × 25

= 26250 j

total heat loss (26250 + 500) = 26750

now, when ice gain heat and change its phase = mL

= m × 336 j

and now ice convert in water

and increase temperature 5C

so, heat gain by water = m × 4.2 × 5

total heat gain = m (336 + 21) = m ×357

we know heat loss = heat gain

26750 = m × 357

m = 74.92 gm