On a winter day the temperature drops from -5 degree C to -15 degree C overnight. If a pan sitting outside contains 0.40kg of ice, how much heat is removed from the ice for the temperature change? (specific heat (ice) = .5kcal/kg) Is it 02.0kcal?
M*C*delta T = 0.4*0.5*10 = 2.0 kCal
You are right!
Well, let's calculate it using some frosty humor, shall we? The temperature drop from -5°C to -15°C overnight sounds like someone forgot to pay their heating bill! Brrrrr! Now, back to the question. To find out how much heat is removed from the ice, we need to use the specific heat of ice, which is 0.5 kcal/kg.
Given that the pan contains 0.40 kg of ice, we can multiply it by the specific heat to get the amount of heat removed. So, 0.40 kg * 0.5 kcal/kg = 0.20 kcal.
So yes, it seems like the amount of heat removed from the ice for the temperature change is indeed 0.20 kcal. Stay cool!
To calculate the amount of heat removed from the ice, we can use the formula:
Q = m * ΔT * specific heat
Where:
Q is the amount of heat removed (in kcal)
m is the mass of the ice (in kg)
ΔT is the change in temperature (in degrees Celsius)
specific heat is the specific heat of ice (in kcal/kg)
Given:
m = 0.40 kg
ΔT = -15°C - (-5°C) = -15°C + 5°C = -10°C
specific heat of ice = 0.5 kcal/kg
Using the formula, we can calculate:
Q = 0.40 kg * -10°C * 0.5 kcal/kg
Q = -2 kcal
The amount of heat removed from the ice is -2 kcal. So, the correct answer is -2 kcal, not 02.0 kcal.
To calculate the amount of heat removed from the ice due to the temperature change, we can use the equation:
Q = m * ΔT * C
Where Q represents the amount of heat, m is the mass of the ice, ΔT is the change in temperature, and C is the specific heat capacity of ice.
Given data:
Mass of ice (m) = 0.40 kg
Change in temperature (ΔT) = -15°C - (-5°C) = -10°C (note the negative sign indicates a decrease in temperature)
Specific heat capacity of ice (C) = 0.5 kcal/kg
Now we can substitute these values into the equation to find the amount of heat (Q):
Q = 0.40 kg * (-10°C) * 0.5 kcal/kg
Q = -2 kcal
Therefore, the amount of heat removed from the ice due to the temperature change is -2 kcal. Note that the negative sign indicates that heat is being removed, as the temperature is decreasing.