How much heat transfer is necessary to raise the temperature of a 13.6 kg piece of ice from −20.0ºC to 130ºC? specific heat capacity of water is 4184 J/kg C and specific latent heat of fusion is 334000 J/kg. 8535360 J
4542400 J
458720900 J
13077760 J

A: 8535360 J

Explanation:

We are given;

mass; m = 13.6 kg

Initial temperature; T1 = -20° C

Final temperature; T2 = 130° C

Specific heat capacity; c = 4184 J/kg.°C

Specific latent heat of fusion; L = 334000 J/kg.

Since we are dealing with change in temperature, then we will use the heat energy formula for specific heat capacity.

Q = mcΔt

Thus;

Q = 13.6 × 4184 × (130 - (-20))

Q = 13.6 × 4184 × 150

Q = 8535360 J

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west