A 43.70 g sample of a substance is initially at 24.7 °C. After absorbing 1757 J of heat, the temperature of the substance is 120.5 °C. What is the specific heat (c) of the substance?
q=m*c*deltaT
q=heat energy
m=mass
c=specific heat
deltaT= T_final - T_initial
To find the specific heat (c) of the substance, we can use the formula:
Q = mcΔT
Where:
- Q is the amount of heat absorbed (in Joules)
- m is the mass of the substance (in grams)
- c is the specific heat of the substance (in J/g°C)
- ΔT (delta T) is the change in temperature (in °C)
We are given the following information:
- Mass (m) = 43.70 g
- Initial temperature (T1) = 24.7 °C
- Final temperature (T2) = 120.5 °C
- Amount of heat absorbed (Q) = 1757 J
First, we need to calculate the change in temperature (ΔT):
ΔT = T2 - T1
= 120.5 °C - 24.7 °C
= 95.8 °C
Now, we can rearrange the formula Q = mcΔT to solve for c:
c = Q / (m * ΔT)
Substituting the given values into the equation:
c = 1757 J / (43.70 g * 95.8 °C)
Calculating the specific heat:
c = 0.0381 J/g°C
Therefore, the specific heat (c) of the substance is approximately 0.0381 J/g°C.