Question

Ten grams of steam absorbs 60.0J of heat. What is the temperature increase of the steam? Specific heat of steam: 1.87J/gC

Answer 1

The temperature increase of the steam is 3.21 °C.

Step-by-step explanation:

To calculate the temperature increase of the steam, we can use the formula Q = mcΔT, where Q is the heat absorbed, m is the mass of the steam, c is the specific heat of steam, and ΔT is the change in temperature. Rearranging the formula, we have ΔT = Q / (mc).

Given that 10 grams of steam absorbs 60.0 J of heat and the specific heat of steam is 1.87 J/g°C, we can substitute these values into the formula: ΔT = 60.0 J / (10 g * 1.87 J/g°C) = 3.21 °C.

Therefore, the temperature increase of the steam is 3.21 °C.

Answer 2

In this question, we have a situation in which we have to use the Calorimetry formula, which is how much heat was released or absorbed (in Joules), after we had a change in temperature of a compound. The formula for this Calorimetry question is:

Q = mcΔT

Where:

Q = is energy as Heat, 60.0 J

m = mass in grams, 10 grams

c = is the specific heat capacity, 1.87 J/g°C

ΔT = the change in temperature, calculated as Final Temperature - Initial T

Now we add these values into the formula:

60 = 10 * 1.87 * ΔT

60 = 18.7ΔT

ΔT = 60/18.7

ΔT = 3.2°C

The temperature had a 3.2°C of increase