Question

How much heat is required to take a 150 g sample of water from 10.0 ℃ to 95.0 ℃? cs,water = 4.184 J/g*℃

Answer 1

Answer:

$\boxed {\boxed {\sf 53, 346 \ Joules}}$

Explanation:

We are given the specific heat and change in temperature, so we should use this heat formula:

$q=m C \Delta T$

where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

We know the mass is 150 grams. The specific heat of water is 4.184 J/g °C.

Let's find the change in temperature.

Subtract the initial temperature from the final temperature.

• ΔT= final temp - initial temp
• final= 95.0 °C and  initial= 10.0 °C
• ΔT= 95.0 °C - 10.0 °C= 85.0 °C

Now we know all the values:

$m= 150 \ g \\C= 4.184 J/ g \ \textdegree C \\\Delta T= 85.0 \textdegree C$

Substitute them into the formula.

$q=(150 \ g) (4.184 \ J/g \ \textdegree C)(85.0 \textdegree C )$

Multiply all three numbers together. Note that the grams (g) and degrees Celsius (°C) will cancel out. Joules (J) will be the only remaining unit.

$q=(627.6 \ J/ \textdegree C) ( 85.0 \textdegree C)$

$q=53346 \ J$

53,346 Joules of heat are required.