The specific heat of liquid bromine is 0.226J/g-K and the density is 3.12g/mL, how much heat (J) is required to raise the temperature of 10.0mL of bromine from 25.00 C to 27.30 C?

Question

Noah Aldrich · Accepted Answer

#"16.2 J"#

The idea here is that you need to use bromine's density to determine how many grams of bromine you have in that #"10.0-mL"# sample.

A density of #"3.12 g/mL"# tells you that you get a mass of #"3.12 g"# for every #"100 mL"# of bromine. This means that your sample will have a mass of

#10.0color(red)(cancel(color(black)("mL"))) * "3.12 g"/(1color(red)(cancel(color(black)("mL")))) = "31.2 g"#

Now, the equation that establishes a relationship between heat absorbed and change in temperature looks like this

#color(blue)(q = m * c * DeltaT)" "#, where
#q# - heat absorbed
#m# - the mass of the sample
#c# - the specific heat of the substance
#DeltaT# - the change in temperature, defined as final temperature minus initial temperature
Plug in your values and solve for #q# to get

#q = 31.2color(red)(cancel(color(black)("g"))) * 0.226"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (27.30 - 25.00)color(red)(cancel(color(black)(""^@"C")))#
#q = color(green)("16.2 J")#

Natalie Anton · Answer

undefined To calculate the heat required, you can use the formula: Q = m * c * ΔT, where Q is the heat, m is the mass, c is the specific heat, and ΔT is the change in temperature. First, you need to find the mass of 10.0 mL of bromine using its density. Then, you can calculate the heat using the given specific heat and the change in temperature.

Given:
Specific heat of bromine (c) = 0.226 J/g-K
Density of bromine = 3.12 g/mL
Volume of bromine (V) = 10.0 mL
Initial temperature (T1) = 25.00°C
Final temperature (T2) = 27.30°C

1. Calculate mass (m) using density:
mass = density * volume
mass = 3.12 g/mL * 10.0 mL
mass = 31.2 g

2. Calculate the change in temperature (ΔT):
ΔT = T2 - T1
ΔT = 27.30°C - 25.00°C
ΔT = 2.30°C

3. Use the formula to calculate heat (Q):
Q = m * c * ΔT
Q = 31.2 g * 0.226 J/g-K * 2.30°C
Q ≈ 16.83 J

Therefore, approximately 16.83 Joules of heat is required to raise the temperature of 10.0 mL of bromine from 25.00°C to 27.30°C.