you take 250.0 g of ice from the refrigerator at -8.00 degrees celsius. how much heat does it take to warm it to room temperature, 20.0 degrees celsius?
add the following:
heat ice to zero
melt ice at zero
heat water from zero to 20C
if you get really lazy you can always use calories (1 small calorie=the energy to increase 1 gram of water by 1 degree Celsius)
Think why that wont work, Jay: We are dealing with ice, then ice melting at constant temperature, then for the lazy, heating water.
To determine how much heat is needed to warm the ice to room temperature, you can use the specific heat formula:
Heat = mass × specific heat capacity × change in temperature
First, let's find the specific heat capacity of ice. The specific heat capacity of ice is typically around 2.09 J/g°C.
Next, let's calculate the change in temperature. The initial temperature is -8.00°C, and the final temperature is 20.0°C. Therefore, the change in temperature is:
Change in temperature = final temperature - initial temperature
Change in temperature = 20.0°C - (-8.00°C)
Change in temperature = 28.0°C
Now, we can calculate the heat required using the formula:
Heat = mass × specific heat capacity × change in temperature
Given:
Mass of ice = 250.0 g
Specific heat capacity of ice = 2.09 J/g°C
Change in temperature = 28.0°C
Heat = 250.0 g × 2.09 J/g°C × 28.0°C
Heat = 14,630 J (or 14.63 kJ)
Therefore, it will take approximately 14,630 Joules (or 14.63 kilojoules) of heat to warm 250.0 grams of ice from -8.00°C to 20.0°C.