Question 6.5: Relating Heat and Specific Heat Calculate the heat absorbed ...

Relating Heat and Specific Heat

Calculate the heat absorbed by 15.0 g of water to raise its temperature from 20.0°C to 50.0°C (at constant pressure). The specific heat of water is 4.18 J/(g • °C).

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You substitute into the equation

q=s \times m \times \Delta t

The temperature change is

\Delta t=t_f-t_i=50.0^{\circ} C -20.0^{\circ} C =+30.0^{\circ} C

Therefore,

q=4.18  J /\left( g \cdot{ }^{\circ} C \right) \times 15.0  g \times\left(+30.0^{\circ} C \right)=1.88 \times 10^3  J

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