Ice+lab+(Temperature+vs.+Time)

ºQuestion: 6.2kg of water at -20ºC is heated to 115ºC. How much energy did I add? Given: mass = 6.2 kg T1 = -20ºC T2 = 0ºC Change in T: 0 +20 = 20ºC Cice = 2114J/kgºC Q=mc⁴T =(6.2kg)(2114J/kgºC)(20ºC) =262,136J Therefore, there is 262,136J of heat. || Given: mass= 6.2kg Lf = 334,000J/Kg Q = m*Lf =(6.2kg)(334,000J/kg) =2,070,800J Therefore, there is 2,070,800J of heat. || Given: mass = 6.2kg T1 = 0ºC T2 = 100ºC ⁴T = 100- 0 = 100ºC Cwater = 4200J/kgºC Q= mc⁴T =(6.2kg)(4200J/kgºC)(100ºC) =2,604,000J Therefore, there is 2,604,000J of heat || Given: mass: 6.2kg Lvapour = 2258,000J/kg Q=mLvapour =(6.2kg)(2258,000J/kg) =13,999,600J Therefore, there is 13,999,600J of heat || Given: mass: 6.2kg T1 + 100ºC T2 =115ºC ⁴T= 115 -100 = 15ºC Cvapour = 2108J/kgºC Q = mc⁴T =(6.2kg)(2108J/kgºC)(15ºC) = 196,044J Therefore, there is 196,044J of heat || The total amount of energy added is 262,136J + 2,070,800J+ 2,604,000J + 13,999,600J + 196,044J =19,132,580J or 19,132.6KJ
 * Step 1: Find Q for ice stage:
 * Step 2: Find Q for melting stage:
 * Step 3: Find Q for liquid water stage:
 * Step 4: Find Q for boiling water stage:
 * Step 5: Find Q for water vapour stage: