How much heat, in joules and calories, is needed to heat up 295 mL of water from 26.6 °C to 85.1 °C? Assume the density of water to be 1.00 g/mL.

heat=mass*specificheat*changintemp

= 295ml*1g/ml*1cal/gram*(85.1-26.6)

that give it to you in calories.

change the units of specific heat if you want it in joules.

To calculate the amount of heat required to heat up the water, we can use the equation:

Q = m * c * ΔT

Where:
Q is the amount of heat in joules (J).
m is the mass of water in grams (g).
c is the specific heat capacity of water, which is approximately 4.18 J/g°C.
ΔT is the change in temperature in °C.

First, let's calculate the mass of the water in grams. We are given the density of water, which is 1.00 g/mL, and the volume of water, which is 295 mL. The mass can be obtained using the equation:

mass = volume * density

mass = 295 mL * 1.00 g/mL

mass = 295 g

Next, let's calculate the change in temperature, ΔT. It is the difference between the final temperature (85.1°C) and the initial temperature (26.6°C):

ΔT = final temperature - initial temperature

ΔT = 85.1°C - 26.6°C

ΔT = 58.5°C

Now we can substitute the values into the equation to calculate the amount of heat, Q:

Q = m * c * ΔT

Q = 295 g * 4.18 J/g°C * 58.5°C

Q ≈ 71134.25 J

To convert this value to calories, we can use the conversion factor:

1 calorie = 4.184 J

So, to convert from joules to calories, divide the value of Q by the conversion factor:

Q in calories = 71134.25 J / 4.184 J/cal

Q in calories ≈ 17026.70 cal

Therefore, the amount of heat needed to heat up 295 mL of water from 26.6°C to 85.1°C is approximately 71134.25 joules (J) and 17026.70 calories (cal).