Assume biological substances are 90 percent water and the density of water is 1.0 ✕ 103 kg/m3.

Calculate the equivalent of 20 degrees Celsius in degrees Fahrenheit and Kelvin.

alekssr [168]

Answer:

68 °F, 293.15 K

Explanation:

Fahrenheit, Kelvin and Celsius are the different scales of temperature in which temperature is measured.

Given : T = 20°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

The conversion of T( °C) to T(F) is shown below:

T (°F) = (T (°C) × 9/5) + 32

So,

3 0

3 years ago

If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu

Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

$\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}$ .

Volume of water:

$V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}$ .

Mass of water:

$m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}$ .

Amount of heat that the 15 L water absorbed:

$Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}$ .

What's the mass of the hot steel tool?

The specific heat of carbon steel is $0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}$ .

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

$Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}$ .

$\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}$ .

$m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}$ .

4 0

3 years ago

Food and water are examples of _______ for populations.

Law Incorporation [45]

Needs................

8 0

2 years ago

Read 2 more answers

Neglecting air resistance, if a package is dropped from an airplane flying 100 km/hr at an altitude of 200 meters, how far from

UkoKoshka [18]

Answer:

178 m

Explanation:

4 0

3 years ago

What do you think is the difference between being imaginative in doing science and Doing pseudoscience?

jenyasd209 [6]

I would say that being imaginative in science means finding new ways to find or analyse the data, but which do not make the data false or which do not manipulate the data. Being imaginative in science should not cross the boundary of making the science not reliable

Being imaginative in pseudoscience could mean manipulating the data or even making experiments in such a way as to only obtaining specific data or throwing away certain data.

7 0

2 years ago