The amount of heat given by the water to the block of ice can be calculated by using

where

is the mass of the water

is the specific heat capacity of water

is the variation of temperature of the water.
Using these numbers, we find

This is the amount of heat released by the water, but this is exactly equal to the amount of heat absorbed by the ice, used to melt it into water according to the formula:

where

is the mass of the ice while

is the specific latent heat of fusion of the ice.
Re-arranging this formula and using the heat Q that we found previously, we can calculate the mass of the ice:
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m = mass of the ice added = ?
M = mass of water = 1.90 kg
= specific heat of the water = 4186 J/(kg ⁰C)
= specific heat of the ice = 2000 J/(kg ⁰C)
= latent heat of fusion of ice to water = 3.35 x 10⁵ J/kg
= initial temperature of ice = 0 ⁰C
= initial temperature of water = 79 ⁰C
T = final equilibrium temperature = 8 ⁰C
using conservation of heat
Heat gained by ice = Heat lost by water
m
(T -
) + m
= M
(
- T)
inserting the values
m (4186) (8 - 0) + m (3.35 x 10⁵ ) = (1.90) (4186) (79 - 8)
m = 1.53 kg
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Answer:
200 N
Explanation:
The crowbar is 2 meter, or 200 cm. The effort arm is 160 cm, so the moment arm of the object is 40 cm.
(800 N) (40 cm) = F (160 cm)
F = 200 N