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
The formula you need to use is:
Q = m * c * ΔT
Where:
Q is the amount of heat energy required (in calories)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in cal/g°C)
ΔT is the change in temperature (in °C)
First, you need to calculate the amount of heat required to heat the ice from -10°C to 0°C (the melting point of ice). Since the specific heat capacity of ice is 0.5 cal/g°C, the calculation would be:
Q1 = 50g * 0.5 cal/g°C * (0°C - (-10°C))
Q1 = 50g * 0.5 cal/g°C * 10°C
Q1 = 250 cal
Next, you need to calculate the amount of heat required to melt the ice at 0°C to water at 0°C. The heat required for this phase change is given by:
Q2 = m * L
Where L is the latent heat of fusion (heat required to change a substance from solid to liquid) for water, which is typically 80 cal/g. So:
Q2 = 50g * 80 cal/g
Q2 = 4000 cal
Lastly, you need to calculate the amount of heat required to heat the liquid water from 0°C to 20°C. The specific heat capacity of water is also 0.5 cal/g°C. So:
Q3 = 50g * 0.5 cal/g°C * (20°C - 0°C)
Q3 = 50g * 0.5 cal/g°C * 20°C
Q3 = 500 cal
Now, you can calculate the total amount of heat required by adding up Q1, Q2, and Q3:
Q total = Q1 + Q2 + Q3
Q total = 250 cal + 4000 cal + 500 cal
Q total = 4750 cal
Therefore, the correct answer is:
D) 5250 cal.
To calculate the amount of heat required to heat the ice cube, we need to consider two parts: raising the temperature of the ice from -10°C to 0°C and then melting the ice at 0°C to liquid water at 20°C.
Part 1: Raising the temperature from -10°C to 0°C:
The formula to calculate the amount of heat (q) required is given by: q = mass (m) × specific heat (c) × change in temperature (ΔT).
Here, the mass of the ice cube is 50 grams, the specific heat of ice is 0.5 cal/goC, and the change in temperature is 0°C - (-10°C) = 10°C.
So, for this part, the amount of heat required is:
q1 = 50 g × 0.5 cal/goC × 10°C = 250 cal.
Part 2: Melting the ice at 0°C to liquid water at 20°C:
To melt the ice to liquid water, we need to calculate the heat required for the phase change. The heat of fusion (Hf) for ice is 1 cal/g.
The mass of the ice cube is still 50 grams, and the change in temperature is 20°C - 0°C = 20°C.
So, for this part, the amount of heat required is:
q2 = 50 g × 1 cal/g × 20°C = 1000 cal.
To find the total amount of heat required, we add the heat from both parts:
Total heat = q1 + q2 = 250 cal + 1000 cal = 1250 cal.
Therefore, the correct answer is A) 1250 cal.