A 50 gram ice cube is cooled to -10oC in the freezer. How many calories of heat are required to heat it until it becomes liquid water at 20oC? (The specific heat of ice is 0.5 cal/goC.)
Answer
A) 1250 cal.
B) 4000 cal.
C) 5000 cal.
D) 5250 cal.
Can someone please help me solve this by helping me out with the formula
To determine the number of calories of heat required to heat the ice cube until it becomes liquid water at 20°C, you can use the following formula:
Q = mcΔT
Where:
Q = heat energy in calories
m = mass of the material (ice cube) in grams
c = specific heat capacity of the material (in this case, ice) in cal/g°C
ΔT = change in temperature (final temperature - initial temperature) in °C
Given:
Mass of ice cube (m) = 50 grams
Specific heat of ice (c) = 0.5 cal/g°C
Initial temperature (T₁) = -10°C
Final temperature (T₂) = 20°C
First, we need to calculate the heat energy required to raise the temperature of the ice from -10°C to 0°C:
Q₁ = mcΔT
= (50 g)(0.5 cal/g°C)(0 - (-10)°C)
= (50 g)(0.5 cal/g°C)(10°C)
= 250 cal
Next, we need to calculate the heat energy required to melt the ice at 0°C:
Q₂ = mLf
Where:
Lf = latent heat of fusion of ice, which is 80 cal/g
Q₂ = (50 g)(80 cal/g)
= 4000 cal
Finally, we need to calculate the heat energy required to raise the temperature of the melted ice from 0°C to 20°C:
Q₃ = mcΔT
= (50 g)(1 cal/g°C)(20°C - 0°C)
= (50 g)(1 cal/g°C)(20°C)
= 1000 cal
Now, we can add up all the heat energies required:
Total heat required = Q₁ + Q₂ + Q₃
= 250 cal + 4000 cal + 1000 cal
= 5250 cal
Therefore, the correct answer is D) 5250 cal.