calculate the temperature change that occurs when 229 cal of heat are added to 22g of water

To calculate the temperature change in the given scenario, we need to use the specific heat capacity (also known as the specific heat) of water. The specific heat of water is 1 calorie/gram°C (cal/g°C).

The formula for calculating the heat absorbed or released by a substance is:

Q = m * c * ΔT

Where:
Q = heat absorbed or released (in calories)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in cal/g°C)
ΔT = change in temperature (in °C)

Given:
Q = 229 cal
m = 22 g
c (for water) = 1 cal/g°C

Now, let's rearrange the formula to solve for ΔT:

ΔT = Q / (m * c)

Plugging in the given values:

ΔT = 229 cal / (22 g * 1 cal/g°C)

Simplifying the units:

ΔT = 229 cal / 22 g°C

Using a calculator, we can find:

ΔT ≈ 10.41 °C

Therefore, when 229 cal of heat is added to 22g of water, the temperature change will be approximately 10.41 °C.