A piece of metal with a mass of 5.50 grams changed in temperature from 25.0 to 55.0C when 21.3 joules was absorbed. The specific heat of the metal is
a. 0.256
b. 0.129
c. 3.91
d. 7.75
e. 0.710J g-1 oC-1
desperately need help!
q = mass metal x specific heat metal x delta T.
q = 21.3 J.
Substitute and solve for specific heat.
is it b?
yes
To find the specific heat of the metal, we can use the formula:
q = m * c * ΔT
Where:
q is the heat absorbed or released by the metal
m is the mass of the metal
c is the specific heat of the metal
ΔT is the change in temperature
In this case, we are given:
m = 5.50 grams
ΔT = (55.0°C - 25.0°C) = 30.0°C
q = 21.3 joules
Substituting these values into the formula, we have:
21.3 joules = (5.50 grams) * c * (30.0°C)
Now, solve for c:
c = 21.3 joules / ((5.50 grams) * (30.0°C))
c ≈ 0.129 J g^(-1) °C^(-1)
Therefore, the specific heat of the metal is approximately 0.129 J g^(-1) °C^(-1). So, the correct answer is b. 0.129.