Answer: So the answer would be um I dont
know the answer right now but message me later on and I will help
Explanation: The equation that relates heat (q) to specific heat (c_p) , mass (m), and temperature change (Delta{T}) is shown below.
q=c_p times m times Delta{T} The heat that is either absorbed or released is measured in joules. The mass is measured in grams. The change in temperature is given by Delta{T}= T_f - T_i , where T_f is the final temperature and T_i is the initial temperature. Step 1: List the known quantities and plan the problem .
Known
heat = q = 134 J
mass = m = 15.0 g
Delta{text{T}} = 62.7^circ text{C} - 24.0^circ text{C} = 38.7^circ text{C}
Unknown
c_p text{of cadmium}= ? text{J}/ text{g}^circ text{C}
The specific heat equation can be rearranged to solve for the specific heat.
Step 2: Solve .
c_p=frac{q}{m times Delta{T}}=frac{134 text{ J}}{15.0 text{ g} times 38.7^circ text{C}}=0.231 text{ J/g}^circ text{C}
Step 3: Think about your result .
The specific heat of cadmium, a metal, is fairly close to the specific heats of other metals. The result has three significant figures.
Since most specific heats are known, they can be used to determine the final temperature attained by a substance when it is either heated or cooled. Suppose that a 60.0 g sample of water at 23.52°C was cooled by the removal of 813 J of heat. The change in temperature can be calculated using the specific heat equation.
Delta{T}=frac{q}{c_p times m}=frac{813 text{ J}}{4.18 text{ J/g}^circ text{C} times 60.0 text{ g}}=3.24^circ text{C}
Since the water was being cooled, the temperature decreases. The final temperature is:
T_f=23.52^circ text{C} - 3.24^circ text{C}=20.28^circ text{C}
