A 86.3mL sample of water is initially at 17.4 degC. How many kilojoules (kJ) of heat must be added to raise the temperature of the water to 31.8? The specific heat capacity of water is 4.184 J/g.degC.

How do I set this up to find the answer. Really confused,we haven't been taught this.

heat=masswater*c*deltatemp

= 86.3g*4.184J/gC*(31.8-17.4)

of course, this assumes you know the density of water to be 1g/ml

To solve this problem, we can use the equation:

q = m * c * △T

Where:
q is the amount of heat transferred (in Joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/g°C), and
△T is the change in temperature (in °C).

First, we need to convert the volume of water (86.3 mL) to grams using the density of water, which is 1 g/mL. So, 86.3 mL of water is equal to 86.3 g.

Next, we need to calculate the change in temperature (△T). It is given that the initial temperature is 17.4°C, and we need to raise it to 31.8°C. Thus, the change in temperature is:

△T = final temperature - initial temperature
△T = 31.8°C - 17.4°C
△T = 14.4°C

Now, we can substitute the values into the equation:

q = m * c * △T
q = 86.3 g * 4.184 J/g°C * 14.4°C

Calculating this expression will give you the amount of heat transferred in Joules. To convert this value to kilojoules (kJ), divide the result by 1000 (since 1 kJ = 1000 J).