Graduated cylinder & beaker; both are used for science activities while a measuring cup is mostly used for baking.
Answer:
a) C_v = 1.005 KJ/kgK
b) C_v = 1005.000 J/kgC
c) C_v = 0.240 kcal/kgC
d) C_v = 0.240 Btu/lbmF
Explanation:
Given:
- constant-pressure specific heat C_v = 1.005 KJ/kgC
Find C_v in units of:
a) kJ/kg·K
b) J/g·°C
c) kcal/ kg·°C
d) Btu/lbm·°F
Solution:
a) C_v is Specific heat capacity is the quantity of heat needed to raise the temperature per unit mass. Usually, it's the heat in Joules needed to raise the temperature of 1 gram of sample 1 Kelvin or 1 degree Celsius. Hence,
C_v = 1.005 KJ/kgK
b)
C_v = 1.005 KJ/kgC * ( 1000 J / KJ)
C_v = 1005.000 J/kgC
c)
C_v = 1.005 KJ/kgC * ( 0.239006 kcal / KJ)
C_v = 0.240 kcal/kgC
d)
C_v = 1.005 KJ/kgC * ( 0.947817 Btu / KJ) * ( kg / 2.205 lbm)*(Δ1 C / Δ1.8 F)
C_v = 0.240 Btu/lbmF
Answer:
I am a girl want private photos
Explanation:
SELECT CASE WHEN GROUPING(AccountDescription) = 1
THEN '*ALL*' ELSE AccountDescription END AS Account,
CASE WHEN GROUPING(VendorState) = 1
THEN '*ALL*' ELSE VendorState END AS State,
SUM(InvoiceLineItemAmount) AS LineItemSum FROM AP.dbo.GLAccounts
JOIN AP.dbo.InvoiceLineItems ON GLAccounts.AccountNo = InvoiceLineItems.AccountNo
JOIN AP.dbo.Invoices ON InvoiceLineItems.InvoiceID = Invoices.InvoiceID
JOIN AP.dbo.Vendors ON Invoices.VendorID = Vendors.VendorID
GROUP BY AccountDescription, VendorState WITH CUBE