100 g of a liquid is 25 °C and is heated to raise the temperature of the liquid to 50.0 °C. The thermal energy gained by the liquid is 6100 J. The specific heat of the liquid is

Question

a) 0.82 J/g °C
b) 2.44 J/g °C
c) 2.06 J/g °C
d) 2.02 J/g °C
e) 2.00 J/g °C

Answer 1

To find the specific heat of the liquid, we can use the formula:

Q = m * c * ΔT

Where Q is the thermal energy gained, m is the mass of the liquid, c is the specific heat, and ΔT is the change in temperature.

Given:
Q = 6100 J
m = 100 g
ΔT = 50.0 °C - 25 °C = 25 °C

Substituting these values into the formula, we get:

6100 J = 100 g * c * 25 °C

Rearranging the equation to solve for c, we have:

c = 6100 J / (100 g * 25 °C)
c = 2.44 J/g °C

Therefore, the specific heat of the liquid is 2.44 J/g °C.

The correct answer is (b) 2.44 J/g °C.

Answer 2

are you sure?

Answer 3

Apologies for the mistake. Let's recalculate:

Q = m * c * ΔT

Given:
Q = 6100 J
m = 100 g
ΔT = 50.0 °C - 25 °C = 25 °C

Substituting the given values into the formula, we get:

6100 J = 100 g * c * 25 °C

Rearranging the equation to solve for c, we have:

c = 6100 J / (100 g * 25 °C)
c = 2.44 J/g °C

So, the correct answer is indeed (b) 2.44 J/g °C.

I apologize for any confusion caused by the previous incorrect response.