Question

find the amount of heat required to convert 0 degree Celsius ice to 30 degree Celsius water​

Answer 1

Answer:

First we have to covert the 1 gram of ice at zero degree to water at zero degree and then that water temperature has to be raised to 30 degrees by providing energy .

So the latent heat of 1 gram of ice= Q1 = 334 joules .

The specific heat capacity of water = C = 4.186 joule/(gram*K)

So the heat required to raise the temperature of 1 gram of water from 0 to 30 degree Celsius = Q2

Q2 = m *C*(T2-T1)

= 1*4.186*(30)

= 125.58 joules

So the total energy required is

Q = Q1+Q2

= 334+125.58

= 459.58 joules

Answer 2

Answer:

Solution Heat required = Mass of ice [latent heat of

fusion + 8 profit of heat × temp change +

Latent heat of boiling ]

⇒ΔH=1g[L

f

+8(ΔT)+L

v

]

ΔH=1g[80cal/g+1cal/g(100−0)+540calg]

ΔH=[80+100+540]=7200