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: 0.15 g

Step-by-step 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

Answer 2

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=(4.184J)/(g.^0C)`

Let's plug in the values in the formula:

`20.0J=m*(4.184J)/(g.^0C)*32.0^0C`

`20.0J=m*(133.888J)/(g)`

`m=20.0J((1g)/(133.888J))`

m = 0.15 g

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