What quantity of heat is required to convert 200g of ice at 0.00C into water at 50C?
I know that im suppose to use the E=mc(Delta T) but im not sure how...
Didn't Bob Pursley do this for you?
http://www.jiskha.com/display.cgi?id=1405592944
To solve this problem using the equation Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature, we need to follow these steps:
Step 1: Determine the heat required to change the ice from its initial temperature of 0.00°C to its melting point, 0.00°C.
Q1 = mcΔT1
In this case, the mass (m) is given as 200 grams, the specific heat capacity (c) for ice is approximately 2.09 J/g°C, and ΔT1 is the change in temperature from 0.00°C to 0.00°C, which is 0.
Q1 = (200 g) x (2.09 J/g°C) x (0 °C)
Q1 = 0 J
So, no heat is required to change the ice from 0.00°C to 0.00°C.
Step 2: Determine the heat required to melt the ice into water at 0.00°C.
Q2 = mLf
In this case, the mass (m) is still 200 grams, and the latent heat of fusion (Lf) for water is approximately 334 J/g.
Q2 = (200 g) x (334 J/g)
Q2 = 66800 J
So, 66800 J of heat is required to melt the ice into water at 0.00°C.
Step 3: Determine the heat required to raise the temperature of the water from 0.00°C to 50°C.
Q3 = mcΔT3
In this case, the mass (m) is still 200 grams, the specific heat capacity (c) for water is approximately 4.18 J/g°C, and ΔT3 is the change in temperature from 0.00°C to 50.00°C.
Q3 = (200 g) x (4.18 J/g°C) x (50 °C)
Q3 = 41800 J
So, 41800 J of heat is required to raise the temperature of the water from 0.00°C to 50.00°C.
Step 4: Calculate the total heat required.
Total heat required = Q1 + Q2 + Q3
Total heat required = 0 J + 66800 J + 41800 J
Total heat required = 108600 J
Therefore, 108,600 J of heat is required to convert 200g of ice at 0.00°C into water at 50.00°C.