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What else can I help you with?
How do you calculate delta e if q equals 1.78 kJ and w equals -862 J?
863.78
Q plus r equals s so t plus u equals?
v
P-q and q-p are logically equivalent prove?
p --> q and q --> p are not equivalent p --> q and q --> (not)p are equivalent The truth table shows this. pq p --> q q -->(not)p f f t t f t t t t f f f t t t t
Q-2.4 equals 1.8 q equals?
1.8q = q - 2.4 1.8q - q = -2.4 0.8q = -2.4 q = -2.4/0.8 q = -2.9999
What is the truth table for negation p or negation q?
P Q (/P or /Q) T T F T F T F T T F F T
What percent of copper in ore is cost efficient to mine?
Q=MC(DELTA)t
What is the Relationship between mass and temperature?
if q= mc delta T then we know that as the mass increases the heat transferred increases
What is the equation for specific equations?
Q = mc(delta)T Q = quantity of heat energy m = mass c = specific heat capacity different constant for each different substance (delta)T = difference in temperature (subtract high temp - low temp)
If 705g of water is heated 889J how much will its temperature increase?
To find the temperature increase, you can use the formula: ( Q = mc\Delta T ), where ( Q ) is the heat energy transferred, ( m ) is the mass of the substance, ( c ) is the specific heat capacity of the substance, and ( \Delta T ) is the change in temperature. Rearrange the formula to solve for ( \Delta T ) by dividing both sides by ( mc ): ( \Delta T = \frac{Q}{mc} ). Substitute ( Q = 889J ), ( m = 705g = 0.705kg ), ( c = 4.18 J/g°C ), and calculate to find the change in temperature.
How many degrees will J raise the temperature of g of water?
To determine how many degrees J will raise the temperature of g of water, we need to use the specific heat capacity formula: ( Q = mc\Delta T ), where ( Q ) is the heat added (in joules), ( m ) is the mass of the water (in grams), ( c ) is the specific heat capacity of water (approximately 4.18 J/g°C), and ( \Delta T ) is the change in temperature (in °C). Rearranging the formula gives ( \Delta T = \frac{Q}{mc} ). Without specific values for Q and g, we cannot calculate the exact change in temperature.
What is the formula to find specific heat of water Q?
The formula to find the specific heat of water ( Q ) is: ( Q = mc\Delta T ), where (m) is the mass of the water, (c) is the specific heat capacity of water, and ( \Delta T ) is the change in temperature of the water.
Q equals m plus x t plus c In this equation what does Q t and c represent?
In the equation Q equals m plus x t plus c, Q represents the total quantity or value being measured or calculated. t represents the variable or time period being observed or measured. c represents the constant term or the y-intercept, which is the value of Q when t equals zero.
How many degrees will 340 J raise the temperature of 6.8g of water?
To find the temperature change, we need to use the formula: ( Q = mc\Delta T ), where ( Q ) is the heat, ( m ) is the mass, ( c ) is the specific heat capacity of water (4.18 J/g°C), and ( \Delta T ) is the temperature change. Substituting the values, we get: ( 340 = 6.8 \times 4.18 \times \Delta T ). Solving for ( \Delta T ), we find that the temperature will rise by approximately 12.75 degrees Celsius.
If 57j of heat are added to an aluminum can with a mass of 17.1g what is the temperature change?
To find the temperature change, use the formula ( q = mc\Delta T ), where ( q ) is the heat added, ( m ) is the mass, ( c ) is the specific heat capacity, and ( \Delta T ) is the temperature change. The specific heat capacity of aluminum is approximately ( 0.897 , \text{J/g}^\circ\text{C} ). Rearranging the formula gives ( \Delta T = \frac{q}{mc} ). Plugging in the values: [ \Delta T = \frac{57 , \text{J}}{17.1 , \text{g} \times 0.897 , \text{J/g}^\circ\text{C}} \approx 3.66^\circ\text{C} ] Thus, the temperature change is approximately ( 3.66^\circ\text{C} ).
When does q equal delta h in a chemical reaction?
Q equals delta H in a chemical reaction when the reaction is at constant pressure and the temperature remains constant.
Suppose you want to heat 40g of water by 20 degrees Celsius how many joules of heat are required?
Use the equation q=mc(delta t) (that is, heat equals mass times specific heat times the change in temperature) to answer the question. The specific heat of water is 4.186 Joules per gram-Celsius. Therefore, q=(40)(4.186)(20), which equals 3348.8 Joules of heat (or approximately 3.35 kiloJoules of heat).
How do you calculate Delta E if q equals 1.40 kJ and w equals -657 J?
658.4
