Question

we have 15g ice with temperature -10 we want to convert to water 18oc

Answer 1

Answer -

Given,

Mass of Ice = 15 g

Temperature of Ice = -10 ^$\degree$ C

Temperature of water = 18 ^$\degree$ C

$\Delta$H_fus of water at 0 ^$\degree$ C = 334 J/g

Specific heat of Ice = 2.108 J°C⁻¹g⁻¹

Specific heat of Water = 4.184 J°C⁻¹g⁻¹

Heat required = ?

For this process,

First, heat (Q1) required to bring temperature form -10 ^$\degree$ C to 0 ^$\degree$ C

Second, heat (Q2) required to melt ICE to WATER at 0 ^$\degree$ C

Third, heat (Q3) required to increase the temperature form 0 ^$\degree$ C to 18 ^$\degree$ C

Total Heat required = Q1 + Q2 + Q3

We know that,

Heat = m * c * $\Delta$ T

where,

m = mass

c = specific heat

$\Delta$T = T_final - T_initial

Now,

Heat = m * $\Delta$ H_fus

where,

m = Mass

$\Delta$H_fus = Heat of Fusion

Q1 = m * c * $\Delta$ T

Q1 = 15 g * 2.108 J°C⁻¹g⁻¹ * [0 ^$\degree$ C - (-10 ^$\degree$ C)]

Q1 = 316.2 J

Q2 = m * $\Delta$ H_fus

Q2 = 15 g * 334 J/g

Q2 = 5010 J

Q3 = m * c * $\Delta$ T

Q3 = 15 g * 4.184 J°C⁻¹g⁻¹ * [18 ^$\degree$ C - (0 ^$\degree$ C)]

Q3 = 1129.68 J

Total Heat required = 316.2 J + 5010 J​​​​​​​ + 1129.68 J

Total Heat required = 6455.88 J or 6.456 kJ [ANSWER]