If it takes 600 calories of heat to raise 75ml of water to a final temp of 96 degrees C. What is the initial temp.
How do I set this problem up if I don't know my change in temp?
To solve this problem, you can use the equation:
q = m * c * ΔT
Where:
- q is the heat energy in calories
- m is the mass of the substance (in this case, water) in grams
- c is the specific heat capacity of water (approximately 1 calorie/gram °C)
- ΔT is the change in temperature in degrees Celsius
Given information:
- Heat energy (q) = 600 calories
- Mass (m) = 75 grams
- Specific heat capacity (c) = 1 calorie/gram °C
Since we are trying to find the initial temperature, and we do not know the change in temperature (ΔT), we can rearrange the equation to solve for the initial temperature (T_initial).
Rearranging the equation, we have:
q = m * c * ΔT
600 calories = 75 grams * 1 calorie/gram °C * ΔT
Now, we can solve for ΔT:
ΔT = 600 calories / (75 grams * 1 calorie/gram °C)
ΔT = 8 °C
Now, we can find the initial temperature (T_initial) by using the given final temperature (T_final) of 96 °C and the change in temperature (ΔT):
T_initial = T_final - ΔT
T_initial = 96 °C - 8 °C
T_initial = 88 °C
Therefore, the initial temperature is 88 degrees Celsius.
To solve this problem, you can use the principle of heat transfer, where the heat gained or lost by a substance is equal to the product of its mass, specific heat capacity, and change in temperature.
In this case, you know the amount of heat (600 calories), the mass of water (75 ml), and the final temperature (96 degrees C). However, you do not know the initial temperature.
To set up the problem without knowing the change in temperature, you can assume a variable for the change in temperature, such as ΔT. By doing this, you can express the change in temperature as a formula: final temperature - initial temperature = ΔT.
The formula for heat transfer is: heat = mass x specific heat capacity x change in temperature.
In this case, the mass is given as 75 ml, but it is preferable to use grams since the specific heat capacity for water is usually given in g/calories. To convert from milliliters to grams, you need to know the density of water, which is approximately 1 g/ml. Therefore, the mass of water can be calculated as 75 grams.
The specific heat capacity of water is approximately 1 calorie/gram°C.
The heat equation can now be written as: 600 calories = 75 grams x 1 calorie/gram°C x ΔT.
To isolate ΔT, divide both sides of the equation by (75 grams x 1 calorie/gram°C):
600 calories / (75 grams x 1 calorie/gram°C) = ΔT.
Simplifying the expression gives you: 8°C = ΔT.
So, the change in temperature, ΔT, is 8 degrees Celsius.
Now, to find the initial temperature, subtract the change in temperature from the final temperature:
Initial temperature = Final temperature - ΔT = 96°C - 8°C = 88°C.
Therefore, the initial temperature of the water is 88 degrees Celsius.