It took 400 calories of heat to raise 50 mL of water from 20 degrees Celsius to its final temperature. What was that temperature?

Ok I understand the equation q=mass x SH x change in temp. I am lost on how to set it up to find my final temp. Please help.

To solve this problem, we can use the equation q = mass × specific heat × change in temperature. In this case, the heat (q) is given as 400 calories, the mass is 50 mL of water, and the initial temperature is 20 degrees Celsius. We need to find the final temperature.

The specific heat capacity (SH) of water is approximately 1 calorie/gram °C or 4.184 Joules/gram °C. However, you have given the mass in mL, so we need to convert it to grams.

First, we need to convert the volume of water (50 mL) to grams using the density of water. The density of water at room temperature is close to 1 g/mL, so 50 mL of water will also be approximately 50 grams.

Let's use the equation q = mass × specific heat × change in temperature and solve for the final temperature (Tf).

q = 400 calories = mass (in grams) × specific heat × change in temperature

Substituting the known values:
400 calories = 50 grams × 1 calorie/gram °C × (Tf - 20 °C)

We can now solve for the final temperature (Tf):

400 calories = 50 grams × (Tf - 20 °C)

Divide both sides by 50 grams:

400 calories / 50 grams = Tf - 20 °C

8 °C = Tf - 20 °C

Add 20 °C to both sides:

8 °C + 20 °C = Tf - 20 °C + 20 °C

28 °C = Tf

Therefore, the final temperature (Tf) is 28 degrees Celsius.