calculate the specific heat of 200g of substance whose temperature is 30 degree celcius if it is mixed with 200g of water heated in a 300g iron container at 100 degree celcius. the final temperature of the mixture is 98 degree celcius.
see other post.
To calculate the specific heat of a substance, we can use the formula:
q = m * c * ΔT,
where q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.
In this case, we need to find the specific heat of the substance.
Given:
Mass of substance (m1) = 200g
Initial temperature of substance (T1) = 30°C
Mass of water (m2) = 200g
Initial temperature of water (T2) = 100°C
Final temperature of the mixture (Tf) = 98°C
First, let's calculate the heat transferred to the substance (q1) using the formula:
q1 = m1 * c1 * ΔT1,
where ΔT1 = Tf - T1.
ΔT1 = 98°C - 30°C = 68°C.
q1 = 200g * c1 * 68°C.
Next, let's calculate the heat transferred to the water (q2) using the formula:
q2 = m2 * c2 * ΔT2,
where ΔT2 = Tf - T2.
ΔT2 = 98°C - 100°C = -2°C.
q2 = 200g * c2 * (-2°C).
Since the heat gained by the substance is equal to the heat lost by the water in this case, we have:
q1 = -q2.
Therefore, we can equate the two equations:
200g * c1 * 68°C = 200g * c2 * (-2°C).
Dividing both sides by 200g and canceling out the mass:
c1 * 68°C = c2 * (-2°C).
Now, let's solve for c1, which is the specific heat of the substance:
c1 = (c2 * -2°C) / 68°C.
Using the information that the specific heat of water (c2) is approximately 4.18 J/g°C, we can substitute the value into the equation:
c1 = (4.18 J/g°C * -2°C) / 68°C.
c1 = -0.123 J/g°C.
Therefore, the specific heat of the substance is approximately -0.123 J/g°C.