How much heat energy (Joules) is required to to raise the temperature of 120.0 g of water from -90 C to -5 C?

Apr 24, 2018

$\text{21.5 kJ}$

Explanation:

Use this equation

$\text{Q = mC"Δ"T}$

where

• $\text{Q =}$ Heat
• $\text{m =}$ Mass of sample
• $\text{C =}$ Specific heat of sample ($\text{2.108 J/g°C}$ for ice)
• $\text{ΔT =}$ Change in temperature

$\text{Q" = 120 cancel"g" × "2.108 J"/(cancel"g" cancel"°C") × [-5 - (-90)] cancel"°C" = "21501.6 J" ≈ "21.5 kJ}$