How much heat(energy) in calories will it take to warm 25 g of ice at -10 degrees C to liquid water at 40 degrees C?

q1 = heat to warm ice from -10 to zero degrees C (without melting it).

q1 = mass ice x specific heat ice x delta T (delta T is 10).

q2 = heat to melt ice at zero degrees C.
q2 = mass ice x heat fusion.

q3 = heat to raise temperature of water zero C to +40.
q3 = mass water x specific heat water x delta T.
Total heat energy is q1 + q2 + q3.
You will need to look up specific heat ice, specific heat water and heat fusion of water.Post your work if you get stuck.
q2 =

To calculate the amount of heat required to warm ice to liquid water at a different temperature, we need to consider two steps: heating the ice to its melting point, and then heating the resulting water from its melting point to the desired final temperature.

Let's break down the calculation into two parts:

1. Heating the ice to its melting point:
The specific heat capacity of ice is 0.5 calories per gram per degree Celsius (cal/g°C). Therefore, the amount of heat required to raise the temperature of 25 grams of ice from -10°C to 0°C can be calculated as follows:

Heat = mass * specific heat capacity * change in temperature

Heat = 25 g * 0.5 cal/g°C * (0°C - (-10°C))

Heat = 25 g * 0.5 cal/g°C * 10°C

Heat = 125 calories

2. Heating the resulting water from its melting point (0°C) to the final temperature (40°C):
The specific heat capacity of liquid water is 1 calorie per gram per degree Celsius (cal/g°C). Therefore, the amount of heat required to raise the temperature of the resulting water from 0°C to 40°C can be calculated as follows:

Heat = mass * specific heat capacity * change in temperature

Heat = 25 g * 1 cal/g°C * (40°C - 0°C)

Heat = 25 g * 1 cal/g°C * 40°C

Heat = 1000 calories

Finally, to find the total amount of heat required, we sum up the heat required in both steps:

Total Heat = Heat to heat ice to its melting point + Heat to heat resulting water to final temperature

Total Heat = 125 calories + 1000 calories

Total Heat = 1125 calories

Therefore, it will take 1125 calories of heat energy to warm 25 grams of ice at -10°C to liquid water at 40°C.