When 300. cal of energy is lost from a 125 g object, the temperature decreases from 45.0°C to 40.0°C. What is the specific heat of this object in Cal/g ‘C?

The answer to your question is C = 0.48 cal/g°C

Explanation:

Data

Heat = Q = 300 cal

mass = m = 125 g

temperature 1 = T1 = 45°C

temperature 2 = T2 = 40°C

Specific heat = C = ? cal/g°C

Formula

    H = mC(T2 - T1)

-Solve for Specific heat (C)

    C = H / m(T2 - T1)

-Substitution

    C = 300 / 125(40 - 45)

-Simplification

    C = -300 / 125(-5)

    C = -300 / -625

-Result

    C = 0.48 cal/g°C

these nuts

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.