Respuesta :
Answer:
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.
Answer:
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