$\text{Dane:}$  

$\text{m}=2\text{kg}$  

${\text{t}}_1={30}^{\circ }\text{C}$   

${\text{t}}_2={800}^{\circ }\text{C}$  

$\text{Q}=198660\text{J}$  

$\text{Szukane:}$  

${\text{c}}_{\text{w}}=\text{?}$  

Obliczamy ciepło właściwe substancji:

${\text{c}}_{\text{w}}=\frac{\text{Q}}{\text{m}\cdot \Delta \text{T}}=\frac{\text{Q}}{\text{m}\cdot \left({\text{t}}_2-{\text{t}}_1\right)}=\frac{198660\text{J}}{2\text{kg}\cdot \left({800}^{\circ }\text{C}-{30}^{\circ }\text{C}\right)}$  

${\text{c}}_{\text{w}}=129\frac{\text{J}}{\text{kg}{\cdot }^{\circ }\text{C}}$  

Odpowiedź: D. 129 ^J/_{(kg ∙ ^oC)} - złoto