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

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

Step-by-step 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)} * 124.9\; {\rm g} * (33.5 - 22.3)\; {\rm K} \\ &= 4.18 \; {\rm J \cdot g^(-1)\cdot K^(-1)} * 124.9\; {\rm g} * 11.2 \; {\rm K} \\ &\approx 5.85 * 10^(3)\; {\rm J}\end{aligned}`.