What mass of water at 15 degrees celcius can be cooled 1 degree celcius by heat necessary to melt 185 grams of ice at 0 degrees celcius?

1052.944 g

Step-by-step explanation:

Given:

Initial temperature of water = 15° C

Final temperature of water = 1° C

Mass of ice = 185 grams

Now,

Heat of fusion of ice = 333.55 J/g

Thus,

The heat required to melt ice = Mass of ice × Heat of fusion

or

The heat required to melt ice = 185 × 333.55 = 61706.75 J

Now,

for water the specific heat capacity= 4.186 J/g.°C

Heat provided = mass × specific heat capacity × Change in temperature

or

61706.75 = mass × 4.186 × (15 - 1)

or

61706.75 = mass × 58.604

or

mass = 1052.944 g

Hence, the mass that can be heated 1052.944 g

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18

(3x-8) or (3x+8)

step-by-step explanation: