Question

The specific heat of water is 4.186. If 100.0 g of water is cooled from 75.0° to 10.0°C, how much heat is released?

a) 0 J. Heat gets absorbed in this situation
b) 27.209 J
c) 30,000 J
d) -27,209 J​

Answer 1

To calculate the amount of heat released when cooling water, we can use the formula:

Q = m × c × ΔT

where:

Q is the heat energy released or absorbed (in Joules),

m is the mass of the water (in grams),

c is the specific heat capacity of water (in J/g°C),

ΔT is the change in temperature (in °C).

Given:

m = 100.0 g (mass of water)

c = 4.186 J/g°C (specific heat capacity of water)

ΔT = 10.0°C (change in temperature, from 75.0°C to 10.0°C)

Using the formula:

Q = 100.0 g × 4.186 J/g°C × (10.0°C - 75.0°C)

Q = 100.0 g × 4.186 J/g°C × (-65.0°C)

Q = -27187 J

The negative sign indicates that heat is released during the cooling process.

Therefore, the correct answer is:

d) -27,209 J