Question

It takes 4500 J to raise the temperature of 2.3 kg of a liquid. If the specific heat capacity of the liquid is 3900 J/kg°C, what is the change in temperature of the liquid?

Answer 1

$$Q=mc\Delta T,$$ $$\Delta T=\dfrac{Q}{mc},$$ \Delta T=\dfrac{4500\ J}{2.3\ kg\cdot3900\ \dfrac{J}{kg\cdot\!^{\circ}C}}=0.5^{\circ}C.