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

y=mx+

fa tha first one you gonna add y to both sides

it look like this

x-y+y=6+y

x=y+6

therefore your answer is

x =y+6

fa tha second one

you gonna flip tha equation

it gonna look like x-4y=y

then add 4 to both sides

it gonna look like tha

x-4 +4 =y +4

so therefore tha answer is x=y+4

step-by-step explanation: