How many joules of energy are needed to change 37.5g of ice at 0.00 oC to water at 45.0oC?

(Heat of fusion = 335J/g,

Specific heat of liquid water = 4.184J/g. oC,

Heat of vaporization = 2259 J/g

19,623 J or 19.6 kJ of heat is needed to change ice at 0°C to water at 45°C

Explanation:

To calculate the energy needed to change 37.5g of ice at 0°C to water at 45.0°C, we obtain the individual values of energy needed to convert the ice from 0° to water at 0 °C and the value of energy needed to convert the water from 0 °C to water at 45°C and then add the values together.

Heat (q) = mΔHf + mCpΔT

So;

1. heat needed to change from solid to liquid = m ΔHf

q = 37.5 * 335

q = 12,562.5 Joules

2. heat needed to convert the water at 0C to water at 45 C

q = mcΔT

q = 37.5 * 4.184 * (45-0)

q = 37.5 * 4.184 * 45

q = 7,060.5 J

The heat needed to change the ice to water at 45 C = 12, 562.5 + 7.060.5 = 19,623 J or 19.6 kJ of heat.

We need 19620 joules

Explanation:

Step 1: Data given

Mass of ice = 37.5 grams

Temperature of ice = 0.00 °C

Final temperature of water = 45.0°C

(Heat of fusion = 335J/g,

Specific heat of liquid water = 4.184J/g°C

Heat of vaporization = 2259 J/g

Step 2: Calculate the energy needed to melt ice to water at 0°C

Q = m*ΔHfus

Q = 37.5 grams * 335J/g

Q = 12562.5 J = 12.56 kJ

Step 3: Calculate energy needed to heat water from 0 to 45 °C

Q = m*c*ΔT

⇒with Q = the enegy needed to heat water from 0 to 45 °C

⇒with m = the mass of water = 37.5 grams

⇒with ΔT = the change of temperature = 45 °C

⇒with c = the specific heat of water = 4.184 J/g°C

Q = 37.5g * 4.184 J/g°C * 45 °C

Q = 7060.5 J = 7.06 kJ

Step 4: Calculate the total heat needed

Total heat = 12.56 kJ + 7.06 kJ

Total heat = 19.62 kJ = 19620 J

We need 19620 joules