User talk:Kvadrate

So the change in energy of the gold is $$E=c*m*(T_1-T_2)$$ where $$c$$ is the specific heat capacity, $$m$$ is the mass, $$T_1$$ is the initial temperature, and $$T_2$$ is the final temperature.

First we have to make sure that the units are right. There are 453.6 grams in a pound, so converting the 37.5 lbs to grams, we have $$\frac{37.5 lb}{1} * \frac{453.6 g}{1 lb} \approx 17,010 g$$

To convert Fahrenheit to Kelvin, you use

$$T_k=(T_f + 459.67)*\frac{5}{9}$$

so 21.1° F is 267.1 K. Since K and Celsius degrees are the same size, we can use these temperatures in our equation above. Substituting these numbers in, it becomes

$$E=(0.13 J/g*K)*(17,010 g)*(267.1 K - 233 K)\approx75,400 J$$ There are 4.184 Joules in 1 calorie, so this means

$$E=\frac{75,400 J}{1} * \frac{1 calorie}{4.184 J} \approx 18,000 calories$$