1. Joe stands at a party holding his insulated foam cup that has nothing in it but 0.100 kg of ice at 0⁰C. How much heat energy in joules must be gained to raise the temperature of the melting ice to room temperature at 23⁰C?
2. Sherry, whose mass is 50 kg, stays out skiing too long and her body temperature drops by 2.00⁰C. How many calories of heat are lost from Sherry's body? (Specific Heat human body= 3470 J/ kg ⁰C)
Can someone please help me I'm struggling in this class😭
I got question 2 figured out but I still need help with question 1
To answer question 1, we need to calculate the amount of heat energy required to raise the temperature of the ice to room temperature.
The equation we can use to calculate the heat energy is:
Q = m * c * ΔT
Where:
Q is the heat energy (in joules),
m is the mass of the ice (in kilograms),
c is the specific heat capacity of water (which is approximately 4190 J/kg°C),
ΔT is the change in temperature (in °C).
Let's use this formula to calculate the heat energy required to raise the temperature of the ice from 0°C to 23°C.
First, we need to calculate the change in temperature (ΔT):
ΔT = final temperature - initial temperature
= 23°C - 0°C
= 23°C
Next, we calculate the heat energy (Q):
Q = m * c * ΔT
= 0.100 kg * 4190 J/kg°C * 23°C
= 9667 J
Therefore, the amount of heat energy required to raise the temperature of the melting ice to room temperature at 23°C is 9667 joules.
Now, let's move on to question 2.
To answer question 2, we need to calculate the amount of heat lost from Sherry's body using the formula:
Q = m * c * ΔT
Where:
Q is the heat energy (in joules),
m is the mass of Sherry's body (in kilograms),
c is the specific heat capacity of the human body (which is approximately 3470 J/kg°C),
ΔT is the change in temperature (in °C).
Given that Sherry's mass is 50 kg and her body temperature dropped by 2.00°C, we can use these values to calculate the heat loss.
First, we calculate the change in temperature (ΔT):
ΔT = final temperature - initial temperature
= 0°C - (-2.00°C)
= 2.00°C
Next, we calculate the heat energy (Q):
Q = m * c * ΔT
= 50 kg * 3470 J/kg°C * 2.00°C
= 347,000 J
Since 1 calorie is equal to approximately 4.1868 joules, we can convert the heat energy from joules to calories:
347,000 J * (1 cal / 4.1868 J) = 82,875 cal
Therefore, approximately 82,875 calories of heat are lost from Sherry's body.