Question

How many grams of water can be heated from 5.0∘C to 37.0∘C with 20.0J of heat? The specific heat of water is 4.184J/g∘C.

Answer 1

Answer:-  mass of water is 0.15 g.

Solution:- This problem is based on the formula:

$q=mC_s\Delta T$

where, q is the heat energy, m is mass, $C_s$ is specific heat and $\Delta T$ is change in temperature.

$\Delta T$  = final temperature - initial temperature

$\Delta T$  = 37.0 - 5.0 = 32.0 degree C

m = ?

q = 20.0 J

$C_s=\frac{4.184J}{g.^0C}$

Let's plug in the values in the formula:

$20.0J=m*\frac{4.184J}{g.^0C}*32.0^0C$

$20.0J=m*\frac{133.888J}{g}$

$m=20.0J(\frac{1g}{133.888J})$

m = 0.15 g

So, the mass of water for the given problem is 0.15 g.

Answer 2

Answer: 0.15 g

Explanation:

m = q / c×ΔT

m = 20 J / 4.84 J/ g C X (37.0 C - 5.0 C)

= 20 J / 4.184 J / g C X (32.0 C)

= 0.149 g