Question

if a sample of water absorbs 41.0 kj of heat and the temperature of the sample changes from 25.0°c to 56.0°c, what is the mass of the sample in grams? remember that the specific heat capacity of water is 4.184 j/(g·°c). provide your answer below: g

Explanation:

Step1: Convert heat from kJ to J

$$Q = 41.0\space kJ\times1000 = 41000\space J$$

Step2: Calculate temperature change

$$\Delta T=T_{final}-T_{initial}=56.0^{\circ}C - 25.0^{\circ}C=31.0^{\circ}C$$

Step3: Rearrange the specific - heat formula \(Q = mc\Delta T\) to solve for mass \(m\)

The formula \(Q = mc\Delta T\) can be rewritten as \(m=\frac{Q}{c\Delta T}\). Given \(c = 4.184\space\frac{J}{g\cdot^{\circ}C}\), \(Q = 41000\space J\), and \(\Delta T=31.0^{\circ}C\)
$$m=\frac{41000\space J}{4.184\space\frac{J}{g\cdot^{\circ}C}\times31.0^{\circ}C}$$
$$m=\frac{41000}{4.184\times31.0}\space g$$
$$4.184\times31.0 = 129.704$$
$$m=\frac{41000}{129.704}\space g\approx315\space g$$

Answer:

315 g