Question

How much heat (in joules) is required to raise the temperature of 36.0 kg of water from 24∘C to 91∘C ? The specific heat of water is 4186 J/kg.C∘.

Answer 1

Step-by-step explanation:

To calculate the amount of heat required to raise the temperature of a given mass of water, you can use the formula:

Q = m * c * ΔT

Where:

Q = heat energy (in joules)

m = mass (in kilograms)

c = specific heat capacity (in J/(kg·°C))

ΔT = change in temperature (in °C)

Given:

m = 36.0 kg

c (specific heat of water) = 4186 J/(kg·°C)

ΔT (change in temperature) = 91°C - 24°C = 67°C

Now, plug these values into the formula to calculate the heat energy (Q):

Q = 36.0 kg * 4186 J/(kg·°C) * 67°C

Q = 96,530,448 J

So, the amount of heat required to raise the temperature of 36.0 kg of water from 24°C to 91°C is approximately 96,530,448 joules.