JEE MAIN - Physics (2011 - No. 16)
$$100g$$ of water is heated from $${30^ \circ }C$$ to $${50^ \circ }C$$. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is $$4184$$ $$J/kg/K$$):
$$8.4$$ $$kJ$$
$$84$$ $$kJ$$
$$2.1$$ $$kJ$$
$$4.2$$ $$kJ$$
Explicação
$$\Delta U = \Delta Q = mc\Delta T$$
$$ = 100 \times {10^{ - 3}} \times 4184\left( {50 - 30} \right) \approx 8.4\,kJ$$
$$ = 100 \times {10^{ - 3}} \times 4184\left( {50 - 30} \right) \approx 8.4\,kJ$$