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
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
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.
are you sure?
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.
To find the specific heat of a substance, we can use the formula:
Q = m * c * ΔT
Where:
Q is the thermal energy gained by the substance (in Joules),
m is the mass of the substance (in grams),
c is the specific heat of the substance (in J/g °C), and
ΔT is the change in temperature (in °C).
Given:
Q = 6100 J
m = 100 g
ΔT = 50 °C - 25 °C = 25 °C
We can rearrange the formula to solve for c:
c = Q / (m * ΔT)
Plugging in the values we have:
c = 6100 J / (100 g * 25 °C)
c = 6100 J / 2500 g °C
c = 2.44 J/g °C
So, the specific heat of the liquid is 2.44 J/g °C.
Therefore, the correct option is b) 2.44 J/g °C.