On a winter day the temperature drops from –15°C to –25°C overnight. If a pan sitting outside contains 0.20 kg of ice, how much heat is removed from the ice for this temperature change?
Well, that's a pretty cool question! Let me break it down for you.
To calculate the heat removed from the ice, we first need to find the change in temperature. So, the temperature dropped from -15°C to -25°C, which means there was a decrease of 10°C.
Now, we need to know the specific heat capacity of ice. But before that, let me ask you this: Why did the scarecrow break up with the cornstalk?
Because it heard it was going to pop! 🌽💔
Anyway, the specific heat capacity of ice is approximately 2.09 J/g°C or 2090 J/kg°C. Since we have 0.20 kg of ice, we can multiply it by the change in temperature and the specific heat capacity of ice to get the heat removed.
So, heat removed = 0.20 kg * 10°C * 2090 J/kg°C.
Calculating that out, we find the heat removed from the ice is 418 J. That's a chilly amount of heat! ❄️😄
To find the amount of heat removed from the ice for this temperature change, we can use the formula:
Q = m * c * ΔT
Where:
Q = Amount of heat removed (in Joules)
m = Mass of the ice (in kg)
c = Specific heat capacity of ice (in J/kg·°C)
ΔT = Change in temperature (in °C)
First, let's calculate the change in temperature (ΔT):
ΔT = final temperature - initial temperature
ΔT = -25°C - (-15°C)
ΔT = -25°C + 15°C
ΔT = -10°C
Next, we need to determine the specific heat capacity of ice. The specific heat capacity of ice is approximately 2,090 J/kg·°C.
Finally, we can calculate the amount of heat removed (Q):
Q = 0.20 kg * 2,090 J/kg·°C * (-10°C)
Q = -41,800 J
Therefore, the amount of heat removed from the ice for this temperature change is -41,800 J (negative sign indicates heat loss).