Question

Suppose a 20.0 g gold bar at 35.0°C absorbs 70.0 calories of heat energy. Given that the specific heat of gold is 0.0310 cal/g °C, what is the final
temperature of the gold bar?

Answer 1

We know, change in temperature is given by :

$T_2-T_1=\dfrac{q}{mC_p(Gold)}$

Putting all given values, we get :

$T_2-T_1=\dfrac{70\ cal}{20\ g\times 0.0310\ cal/g^o\ C}\\\\T_2-T_1=112.90^oC\\\\T_2-35^oC=112.90^oC\\\\T_2=(112.90+35)^oC\\\\T_2=147.9^oC$

Hence, this is the required solution.