# How much the heat needed to raise the temperature of 700 g of water from 25 c to 90 c?

Apr 20, 2018

$Q = 190372$ J or
$Q \approx 190$ KJ

#### Explanation:

To calculate the total heat needed to raise the temperature of 700g of water from 25 C to 90 C you would use the formula

$Q = m c \Delta T$

Where $Q$ is heat (Joules), $m$ is mass (grams), $c$ is the specific heat of the substance ($\frac{J}{g \cdot C}$), and $\Delta T$ is the change in temperature (Celsius).

Specific heat of water, the energy/heat needed to raise the temperature of 1 gram of water by 1 degree Celsius is 4.184$\frac{J}{g \cdot C}$

$\Delta T$ is just the change in temperature $90 C - 25 C = 65 C$

Mass is 700g

From here you just put in each variable and solve for Q:

$Q = 700 \cdot 4.184 \cdot 65$

$Q = 190372$ J or
$Q \approx 190$ KJ

Apr 20, 2018

Approximately $\text{190.4 kJ}$

#### Explanation:

Use this equation

$\text{Q = mC"Δ"T}$

where

• $\text{Q =}$ Heat
• $\text{m =}$ Mass of sample
• $\text{C =}$ Specific heat of sample ($\text{4.184 J/g°C}$ for water)
• $\text{ΔT =}$ Change in temperature

$\text{Q" = 700 cancel"g" × "4.184 J"/(cancel"g" cancel"°C") × (90 - 25) cancel"°C" = "190372 J" ≈ "190.4 kJ}$