# If the temperature of 1 gram of water changes from 22°C to 27°C, how many calories of heat were involved? How many joules?

Jun 6, 2018

$\text{5 cal, 20.92 J}$

#### Explanation:

Use this equation

$\text{Q = mSΔT}$

Where

• $\text{Q =}$ Heat absorbed/released
• $\text{m =}$ Mass of sample
• $\text{S =}$ Specific heat of sample ($\text{1 cal/g°C}$ or $\text{4.184 J/g°C}$ for water)
• $\text{ΔT =}$ Change in Temperature

$\text{Q" = 1 cancel"g" × 1 "cal"/(cancel"g"cancel"°C") × (27 - 22) cancel"°C" = "5 cal}$

or

$\text{Q" = 1 cancel"g" × 4.184 "J"/(cancel"g"cancel"°C") × (27 - 22) cancel"°C" = "20.92 J}$