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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

${\frac {37.5lb}{1}}*{\frac {453.6g}{1lb}}\approx 17,010g$

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

$E=(0.13J/g*K)*(17,010g)*(267.1K-233K)\approx 75,400J$

There are 4.184 Joules in 1 calorie, so this means

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

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