HOW MANY GRAMS OF ICE AT -5.2 DEGREES CELSIUS CAN BE COMPLETELY CONVERTED TO LIQUID AT 6.4 CELSIUS DEGREES IF THE AVAILABLE HEAT FOR THE PROCESS IS 441800kj,FOR ICE SPECIFIC HEAT IS 2.01
Heatavail= warmingice + melting ice + heating water
441.8Kj=m*cice*(0-(5.2C))+Hm*m+m*cwater*(6.4-0)
solve for m.
To determine the number of grams of ice that can be completely converted to liquid, we need to calculate the heat absorbed by the ice and compare it to the available heat for the process.
The equation needed to calculate the heat absorbed is:
Q = m * ΔT * C
Where:
Q is the heat absorbed (in joules)
m is the mass of the substance (in grams)
ΔT is the change in temperature (in Celsius)
C is the specific heat capacity (in joules per gram per Celsius degree)
First, let's calculate the heat absorbed by the ice:
Q_ice = m_ice * ΔT_ice * C_ice
We are given:
ΔT_ice = 6.4°C - (-5.2°C) = 11.6°C
C_ice = 2.01 J/g°C
Q_ice = 441,800 J
Now we can rearrange the equation to solve for the mass of ice (m_ice):
m_ice = Q_ice / (ΔT_ice * C_ice)
m_ice = 441,800 J / (11.6°C * 2.01 J/g°C)
m_ice ≈ 19,034.92 grams
Therefore, approximately 19,034.92 grams of ice at -5.2 degrees Celsius can be completely converted to liquid at 6.4 degrees Celsius, given the available heat of 441,800 kJ and the specific heat capacity of ice as 2.01 J/g°C.