A 300 Calorie donut is used to heat 5.0 kg of water. If the initial temperature of the water is 20 degrees C what will the final temperature of the water be? Assume no heat loss to the surroundings

300cal=5000gwater*1cal/gramC*deltatemp

solve for delta temp.

Where does the 1cal/gramC come from???

To find the final temperature of the water, we can use the formula for heat transfer:

Q = mcΔT

Where:
Q is the amount of heat transferred,
m is the mass of the water,
c is the specific heat capacity of water,
ΔT is the change in temperature.

First, let's convert the calorie unit to Joules, since the specific heat capacity of water is typically given in Joules per gram-degree Celsius. We know that 1 calorie is equivalent to 4.184 Joules.

So, the energy supplied by the donut is:
Q = 300 calories × 4.184 J/cal = 1252.2 J

The mass of the water is given as 5.0 kg. The specific heat capacity of water, c, is approximately 4.18 J/g°C. Since we have the mass in kg, we need to convert it to grams:
m = 5.0 kg × 1000 g/kg = 5000 g

Substituting all the values into the heat transfer equation, we have:

1252.2 J = 5000 g × 4.18 J/g°C × ΔT

Simplifying the equation:

ΔT = 1252.2 J / (5000 g × 4.18J/g°C)
ΔT = 0.0599°C

Therefore, the change in temperature of the water is 0.0599°C. To find the final temperature, we need to add this change to the initial temperature:

Final temperature = Initial temperature + Change in temperature
Final temperature = 20°C + 0.0599°C

Therefore, the final temperature of the water is approximately 20.06°C.