Question

A breaker holds 124.9 g of water. The initial temperature of the water is 22.3 C how much energy is needed to raise the temperature of the water to 33.5 C? The specific heat of the water is 4.18 j/g* C

Answer 1

Answer:

Approximately $5.85 \times 10^{3}\; {\rm J}$ assuming no energy loss to the surroundings of the water in this beaker.

Explanation:

Let $c$ denote the specific heat of a material. The energy $Q$ required to raise the temperature of $m$ (mass) of this material by $\Delta T$ (change in temperature) is:

$Q = c\, m\, \Delta T$.

In this question, it is given that the specific heat of water is $c = 4.18\; {\rm J \cdot g^{-1}\cdot K^{-1}}$. It is also given that the mass of the water in this beaker is $m = 124.9\: {\rm g}$.

The change in the temperature is:

$\Delta T = (33.5 - 22.3)\; {\rm K} = 11.2\; {\rm K}$.

Assume that there is no heat loss to the surroundings of the water in this beaker. Energy required to achieve this change in temperature would be:

$\begin{aligned}Q &= c\, m\, \Delta T \\ &= 4.18 \; {\rm J \cdot g^{-1}\cdot K^{-1}} \times 124.9\; {\rm g} \times (33.5 - 22.3)\; {\rm K} \\ &= 4.18 \; {\rm J \cdot g^{-1}\cdot K^{-1}} \times 124.9\; {\rm g} \times 11.2 \; {\rm K} \\ &\approx 5.85 \times 10^{3}\; {\rm J}\end{aligned}$.